Mechanics: calculating launch angle of projectile

[roam]

Homework Statement

A ball has travelled a horizontal distance of 95m and landed 9m up in a stand as shown in the diagram attached below. The time of the flight was 4.5s. Calculate the angle at which the ball was hit.

Relevant Equations

$\theta=\arctan(\frac{v_{y}}{v_{x}})$

$v^2_y = v^2_{0y} - 2 g(y-y_0)$

$v$ indicates velocity and $y$ is height.

I am trying to find the launch angle $\theta$ using the relationship:

$\theta=\arctan(\frac{v_{y}}{v_{x}})$

So, ignoring attenuation due to air resistance, we have a constant horizontal velocity:

$v_x = d_x/t = 95m/4.5s \approx 21 m/s$

But what value do we use for $v_y$ which is not constant? It has a positive value until the ball reaches the apex of the trajectory where $v_y=0$. Or is there a simpler way to approach this problem?

Any help is greatly appreciated.

 

[Steve4Physics]

Byut what value do we use for $v_y$ which is not constant? It has a positive value until the ball reaches the apex of the trajectory where $v_y=0$. Or is there a simpler way to approach this problem?

You are on the right track.

Let $v_y$ be the vertical component of the initial velocity, i.e. at the moment of launch.

The vertical motion is independent of the horizontal motion. In the vertical direction, the start position is (say) $y_0 = 0$ and the final position is $y=10m$ (taking upwards as positive).
[Edit: sorry, that should be $y=9m$.]

You are given the time of flight (t) and the vertical acceleration (a=g).

Do you know any (standard) equation relating $y, y_0, v_y$ (initial velocity), $a$ and $t$? (Hint: it’s a quadratic equation as there are two times at which the height is 10m.)
[Edit: sorry, that should be 9m.]

 

[rudransh verma]

Homework Statement:: A ball has traveled a horizontal distance of 95m and landed 9m up in a stand as shown in the diagram attached below. The time of the flight was 4.5s. Calculate the angle at which the ball was hit.
Relevant Equations:: $\theta=\arctan(\frac{v_{y}}{v_{x}})$

$v^2_y = v^2_{0y} - 2 g(y-y_0)$

$v$ indicates velocity and $y$ is height.

But what value do we use for vy which is not constant? It has a positive value until the ball reaches the apex of the trajectory where vy=0. Or is there a simpler way to approach this problem?

First of all what is the initial launch velocity vo and the height the projectile attained?
Know this and we can use 3rd motion eqn.

 

[bob012345]

You are on the right track.

Let $v_y$ be the vertical component of the initial velocity, i.e. at the moment of launch.

The vertical motion is independent of the horizontal motion. In the vertical direction, the start position is (say) $y_0 = 0$ and the final position is $y=10m$ (taking upwards as positive).

You are given the time of flight (t) and the vertical acceleration (a=g).

Do you know any (standard) equation relating $y, y_0, v_y$ (initial velocity), $a$ and $t$? (Hint: it’s a quadratic equation as there are two times at which the height is 10m.)

Where did you get $10m$?

 

[roam]

You are on the right track.

Let $v_y$ be the vertical component of the initial velocity, i.e. at the moment of launch.

The vertical motion is independent of the horizontal motion. In the vertical direction, the start position is (say) $y_0 = 0$ and the final position is $y=10m$ (taking upwards as positive).

You are given the time of flight (t) and the vertical acceleration (a=g).

Do you know any (standard) equation relating $y, y_0, v_y$ (initial velocity), $a$ and $t$? (Hint: it’s a quadratic equation as there are two times at which the height is 10m.)

Hi there,

I'm not sure, do you mean $v^2_y = v_{0y}^2 -2a(y-y_0)$?

In that case, we have $v^2_y = v_{0y}^2 -2 (9.81) \times 9$. That is as far as we could go with this equation.

P.S. The final height of the projectile is 9m.

First of all what is the initial launch velocity vo and the height the projectile attained?
Know this and we can use 3rd motion eqn.

Hi there,

By initial launch velocity do you mean $v_{oy} = 0$, or do you mean the vertical velocity immediately after launch which is nonzero?

The problem is that they didn't give us a single initial launch velocity from which we could calculate $x$ and $y$ components. As I've shown above, we can readily calculate the horizontal component but not the vertical one.

 

[PeroK]

Hi there,

I'm not sure, do you mean $v^2_y = v_{0y}^2 -2a(y-y_0)$?

In that case, we have $v^2_y = v_{0y}^2 -2 (9.81) \times 9$. That is as far as we could go with this equation.

P.S. The final height of the projectile is 9m.

You don't know the vertical component of the final velocity, so that equation isn't very useful.

You do know the final height and time of flight. Can you get the vertical component of the initial velocity from that using a different equation?

 

[PeroK]

The problem is that they didn't give us a single initial launch velocity from which we could calculate $x$ and $y$ components.

If you were given the initial velocity it would a) be too easy and b) you wouldn't need to know the final height of $9m$. It's not a good idea to worry about what you are not given in such problems - instead, you should focus on what you are given!

 

[rudransh verma]

@PeroK how is the final height=9. I think the final height would be the max distance the ball traveled up to where the velocity becomes zero. This height is not given.

 

[rudransh verma]

By initial launch velocity do you mean voy=0,

This. But it’s not zero as far as I know.

 

[PeroK]

@PeroK how is the final height=9. I think the final height would be the max distance the ball traveled up to where the velocity becomes zero. This height is not given.

The final height is $9m$. I'm not sure why anyone would look at that diagram and conclude otherwise.

 

[rudransh verma]

The final height is $9m$. I'm not sure why anyone would look at that diagram and conclude otherwise.

If a ball lands on ground what will be the final height?
I say this because the vertical velocity component is positive then zero then negative in its total journey. So I guess we should be taking the component up to where it’s direction stays same.

 

[berkeman]

If a ball lands on ground what will be the final height?

It does not land on the ground...

Homework Statement:: A ball has traveled a horizontal distance of 95m and landed 9m up in a stand

 

[vela]

@PeroK how is the final height=9. I think the final height would be the max distance the ball traveled up to where the velocity becomes zero. This height is not given.

When $v_y = 0$, the ball is still midflight. Why do you think $y$ at this point is its final height?

 

[Steve4Physics]

Hi @roam. My apologies - in Post #2 I took final height y=10m, but of course it is y=9m.

There is a set of standard 'kinematics equations' used in problems where motion is in 1-dimension and acceleration is constant. There are 4 (or 5 depending on where you are taught) equations.

For motion with constant acceleration in, say, the x-direction, the 5 equations are shown here:
https://www.physicsforums.com/attachments/qepqu-png.155340/
The symbols used in the image are:
$v_0$ = the initial velocity
$v$ = the final velocity
$x_0$ = the initial position (x-coordinate)
$x$ = the final position (x-coordinate)
$a$ = the constant acceleration (in the x-direction)
$t$ = the elapsed time between initial and final states
Of course you can change 'x' to 'y' if considering motion in the y-direction.

There are slight variations in the format and symbols used. (Sometimes the symbols s, u, v, a and t are used and the equations are then referred to as the 'suvat' equations.) I hope you recognise these equations,

Can you see how to use one of the equations to find the initial y-component of velocity?

Remember for a projectile in free-fall, the vertical component of motion is unaffected by the horizontal component of motion.

 

[PAllen]

Can you write an equation for y as a function of t given some (unknown) initial y velocity, and initial y value of zero.? Once you can write this equation, note that you are given a final time value and final value of y. Then you can just solve for the remaining unknown (initial y velocity).

 

[roam]

Hi @PeroK and @Steve4Physics,

Thank you so much for your inputs. I believe the equation to use is:

$y-y_0 = v_{y} t + \frac{1}{2} a t^2 \tag{1}$

Is this right? If so, the initial y velocity becomes:

$9 = v_{y} (4.5) + \frac{1}{2} (9.81) (4.5)^2 \implies v_{y} \approx -20 \ \text{m/s}$

(I wonder if for the gravitation I should have used -9.81 to get a positive value for velocity?)
[EDIT: in that case $v_y = 24 \text{m/s}$]

The horizontal component was:

$v_x = \frac{x}{t} = \frac{95}{4.5}\approx21 \ \text{m/s}$

So using trigonometry:

$\text{tan}^{-1} \left( \frac{v_y}{v_x} \right) = \text{tan}^{-1} \left( \frac{24}{21} \right) \approx 49^{\circ}$

Is this correct?

 

[haruspex]

(I wonder if for the gravitation I should have used -9.81 to get a positive value for velocity?)

For the heights, you appear to have defined up as positive, so you should do the same for all vertical accelerations and displacements.
So what sign should a downward acceleration have?

 

[rudransh verma]

When $v_y = 0$, the ball is still midflight. Why do you think $y$ at this point is its final height?

Because after that the velocity will be negative. We should be taking the final height where the velocity component is in one direction.

 

[PeroK]

$\text{tan}^{-1} \left( \frac{v_y}{v_x} \right) = \text{tan}^{-1} \left( \frac{24}{21} \right) \approx 49^{\circ}$

Yes, that looks right. Note that you could have done: $$x = u_xt, \ \ y = u_y t - \frac 1 2 gt^2$$ and $$\tan \theta = \frac{u_y}{u_x} = \frac{u_yt}{u_xt} = \frac{y + \frac 1 2 gt^2}{x} = \frac{9 + (4.9)(4.5^2)}{95} = 1.14$$
Which confirms your answer.

 

[rudransh verma]

I have a confusion. This is the attached copy of a numerical on the 2nd eqn of motion. What does this suggest about the motion? What are the initial and final velocity, time, and displacement

 

[haruspex]

I have a confusion. This is the attached copy of a numerical on the 2nd eqn of motion. What does this suggest about the motion? What are the initial and final velocity, time, and displacement

I strongly advise working entirely symbolically as far as possible, only plugging in numbers at the end. It has many advantages, one being that it is much easier for others to follow the development.

 

[rudransh verma]

I strongly advise working entirely symbolically as far as possible, only plugging in numbers at the end. It has many advantages, one being that it is much easier for others to follow the development.

It is not a numerical to solve. Compare the two images and what do you think the motion is. I came to understand that when displacement is less than max displacement then it will have two velocities. One positive and negative. The negative velocity and corresponding displacement is what I thought is not possible.
But the eqn is built very nicely to include the velocity after the zero velocity at max height.
The eqn very well calculates the displacement based on the time t passed. First the displacement increases and then it again decreases to zero. The displacement depends on the time t we choose.

 

[vela]

Because after that the velocity will be negative.

So what? There's nothing particularly special about that point in time as far as the kinematic equations are concerned.

We should be taking the final height where the velocity component is in one direction.

You're making a mistake students commonly make where they think that they need to break the flight of the ball up into two pieces, one where it goes up and one where it goes down. It's completely unnecessary and increases the amount of work you have to do to get to the final answer.

Try solving the problem the way that has been suggested by others in the thread and the way you're suggesting. You'll see the two methods will yield the same answer.

 

[kuruman]

The solution to this problem is greatly simplified if one considers the final position of the projectile as the sum of two vectors, one along the initial velocity and one along the acceleration of gravity:$$\vec r_{\!f}=\vec v_0 ~t_{\!f}+\frac{1}{2}\vec g~ t_{\!f}^2.$$The drawing below shows the vector addition. Finding the angle of projection using simple trigonometry on right triangle OAB is trivial when the time of flight $t_{\!f}$, horizontal range $R$ and landing height $h$ are known.

 

[PeroK]

The solution to this problem is greatly simplified if one considers the final position of the projectile as the sum of two vectors, one along the initial velocity and one along the acceleration of gravity:$$\vec r_{\!f}=\vec v_0 ~t_{\!f}+\frac{1}{2}\vec g~ t_{\!f}^2.$$The drawing below shows the vector addition. Finding the angle of projection using simple trigonometry on right triangle OAB is trivial when the time of flight $t_{\!f}$, horizontal range $R$ and landing height $h$ are known.

View attachment 291680

Being careful not to extend the observations beyond introductory levels, but that is potentially a general relativistic approach, where there is no force of gravity, but the ground is accelerating upwards!

The projectile travels inertially in a straight line at constant velocity, while the target starts at height $h$ and accelerates upwards at $g$.

 

[kuruman]

Being careful not to extend the observations beyond introductory levels, but that is potentially a general relativistic approach, where there is no force of gravity, but the ground is accelerating upwards!

The projectile travels inertially in a straight line at constant velocity, while the target starts at height $h$ and accelerates upwards at $g$.

Sure, but I was thinking more along the lines of assembling the two relevant kinematic equations into a single vector equation and representing that graphically. No GR here.

 

[PeroK]

Sure, but I was thinking more along the lines of assembling the two relevant kinematic equations into a single vector equation and representing that graphically. No GR here.

That's true, but it does highlight how the nature of gravity as a ficticious force lurks in introductory projectile kinematics!

 

[neilparker62]

The solution to this problem is greatly simplified if one considers the final position of the projectile as the sum of two vectors, one along the initial velocity and one along the acceleration of gravity:$$\vec r_{\!f}=\vec v_0 ~t_{\!f}+\frac{1}{2}\vec g~ t_{\!f}^2.$$

Or using your vector equation $\vec{v} \times \vec{u} = \vec{g} \times \vec{R}$:

Solve for y: {190/9,-sqrt(y^2-2*9.81*9)} cross {190,y} ={0,-9.81} cross {95,0}

Doesn't even need the 't' value but I guess not really appropriate for this homework question.
Edit: this is wrong - we needed the t-value to calculate $v_x=190/9 \approx 21.11\;ms^{-1}$

 

[neilparker62]

We could also use average velocity to determine $v_y$:
$$v_{y_{av}} \times 4.5 =9 \implies v_{y_{av}}=2\;ms^{-1}$$
Exploiting the fact that $v_{y_{av}}$ will be the instantaneous velocity at t=2.25 s:
$$v_{y_i}-2.25 g = 2 \implies v_{y_i}= 2 + 2.25 g \approx 24.07 \;ms^{-1}$$

 

[roam]

Thank you so much for all the explanations. I have only one more follow up question. In post #16 I used one of the kinematic equations to find the "initial vertical velocity":

$y-y_{0}=v_{y}t+\frac{1}{2}at^{2}$
$9-0=v_{y}(4.5)+\frac{1}{2}(-9.81)(4.5)^{2}$
$\therefore v_{y}\approx\ 24 \text{m/s.}$

Is it possible to use this same value to work out the maximum height $h$ of the projectile (at point B of the diagram)?

I mean, there is another kinematic equation that could be used (given that the vertical velocity at the apex is 0):

$v_{y(\text{final})}^{2}=v_{y(\text{initial})}^{2}+2ah$
$0=24^{2}+2(-9.8)h \implies h \approx 29\ \text{m}$

So, was our previously calculated value for $v_{y}$ appropriate to be substituted as $v_{y(\text{initial})}$ in the second equation for finding $h$?

 

[haruspex]

Thank you so much for all the explanations. I have only one more follow up question. In post #16 I used one of the kinematic equations to find the "initial vertical velocity":

$y-y_{0}=v_{y}t+\frac{1}{2}at^{2}$
$9-0=v_{y}(4.5)+\frac{1}{2}(-9.81)(4.5)^{2}$
$\therefore v_{y}\approx\ 24 \text{m/s.}$

Is it possible to use this same value to work out the maximum height $h$ of the projectile (at point B of the diagram)?

I mean, there is another kinematic equation that could be used (given that the vertical velocity at the apex is 0):

$v_{y(\text{final})}^{2}=v_{y(\text{initial})}^{2}+2ah$
$0=24^{2}+2(-9.8)h \implies h \approx 29\ \text{m}$

So, was our previously calculated value for $v_{y}$ appropriate to be substituted as $v_{y(\text{initial})}$ in the second equation for finding $h$?

Yes, but finding one numerical value then plugging that into a formula to find another can lead to accumulation of errors. Keep a digit or two of extra precision in the first calculation.
Alternatively, it is often better to keep the first result in algebraic form and continue that way to the second result, only plugging in numbers at the final step. That might be awkward here, though.

 

[rudransh verma]

So, was our previously calculated value for vy appropriate to be substituted as vy(initial) in the second equation for finding h?

Of course. Vy will not change. It’s same for all instants in this situation.

 

[neilparker62]

I mean, there is another kinematic equation that could be used (given that the vertical velocity at the apex is 0):

$v_{y(\text{final})}^{2}=v_{y(\text{initial})}^{2}+2ah$
$0=24^{2}+2(-9.8)h \implies h \approx 29\ \text{m}$

So, was our previously calculated value for $v_{y}$ appropriate to be substituted as $v_{y(\text{initial})}$ in the second equation for finding $h$?

Conservation of mechanical energy: $$mgh_{max}=½m{v_{yi}}^2$$

 

[kuruman]

$\dots$
I mean, there is another kinematic equation that could be used (given that the vertical velocity at the apex is 0) $\dots$

You can invent your own equation in terms of the given quantities. Knowing the time of flight makes the task easy. You know that the maximum height occurs at half the time of flight and that the vertical velocity is zero when that happens. Then you can write two equations expressing these ideas:
$$h_{\text{max}}=v_{\text{0y}}\frac{t_{\!f}}{2}-\frac{1}{2}g\left(\frac{t_{\!f}}{2}\right)^2~;~~0=v_{\text{0y}}^2-2gh_{\text{max}}.$$The task before you is to use the second equation to substitute $v_{\text{0y}}$ into the first equation and solve the ensuing quadratic.

On edit: This is bad advice. See post #36.

 

[PeroK]

Is it possible to use this same value to work out the maximum height $h$ of the projectile (at point B of the diagram)?

If you know that it was at height $y = 9m$ at time $t = 4.5s$, then you have:
$$y = v_{0y}t - \frac 1 2 g t^2 \ \Rightarrow \ v_{0y} = \frac 1 {t}(y + \frac 1 2 g t^2)$$And, the maximum height $h$ is given by the equation $$2gh = v_{0y}^2 \ \Rightarrow \ h = \frac{v_{0y}^2}{2g}$$ And, as mentioned above, you can either combine those equations or solve for $v_{0y}$ first.