Projectile motion when only given distance and acceleration

[James1019]

TL;DR Summary: Find horizontal velocity?

I have no idea how to solve the problem, the question only provide distance 16cm(h),3.6cm(v) and acceleration = 0

 

[fresh_42]

I think you are supposed to assume gravity on Earth, $9.81 m^2/s.$

 

[haruspex]

The following comments are copied from another thread, of which I took this thread to be a duplicate. The more appropriate response here was "please post your attempt".

1. You have mixed up vertical and horizontal displacements.
2. The given vertical displacement is from launch to max height, but the given horizontal displacement is for the whole trajectory.
3. Only the horizontal acceleration is zero. You need an equation involving vertical acceleration.

 

[James1019]

1. You have mixed up vertical and horizontal displacements.
2. The given vertical displacement is from launch to max height, but the given horizontal displacement is for the whole trajectory.
3. Only the horizontal acceleration is zero. You need an equation involving vertical acceleration.

Do I get the answer right?
Velocity can be negative right?
I am not sure answer is correct

 

[haruspex]

View attachment 334373
Do I get the answer right?
Velocity can be negative right?
I am not sure answer is correct

Your scratchings are unintelligible.
Please post your attempt by typing it in, per forum rules.
Lay it out in a logical sequence with explanation.
Avoid plugging in numbers until the end; create symbols as necessary, using distinct variables.
E.g. you could write $x_{final}=v_xt_{final}$, where $x_{final}=0.16$m.

This has many advantages, one of which is making it much easier for others to follow your reasoning.

 

[James1019]

Your scratchings are unintelligible.
Please post your attempt by typing it in, per forum rules.
Lay it out in a logical sequence with explanation.
Avoid plugging in numbers until the end; create symbols as necessary, using distinct variables.
E.g. you could write $x_{final}=v_xt_{final}$, where $x_{final}=0.16$m.

This has many advantages, one of which is making it much easier for others to follow your reasoning.

⁹
Is it better ?
Do my answer correct?

 

[haruspex]

View attachment 334376⁹
Is it better ?
Do my answer correct?

It's certainly more readable, but…
1. Your diagram does not match the information. As I pointed out in post #3

The given vertical displacement is from launch to max height, but the given horizontal displacement is for the whole trajectory.

2. Your diagram does not show a right triangle. The ‘hypotenuse' is curved. So you cannot write that equation for $\tan(\theta)$.
3. In your equation "s=", you have the vertical acceleration in the expression on the right but for s you substituted a horizontal displacement. Your starting equations should involve only horizontal motion and time or only vertical motion and time.

Start by finding the initial vertical velocity from:

• max height
• vertical velocity at max height
• vertical acceleration

 

[James1019]

It's certainly more readable, but…
1. Your diagram does not match the information. As I pointed out in post #3

2. Your diagram does not show a right triangle. The ‘hypotenuse' is curved. So you cannot write that equation for $\tan(\theta)$.
3. In your equation "s=", you have the vertical acceleration in the expression on the right but for s you substituted a horizontal displacement. Your starting equations should involve only horizontal motion and time or only vertical motion and time.

Start by finding the initial vertical velocity from:

• max height
• vertical velocity at max height
• vertical acceleration

But my question is I cannot find the max hieght without angle
Do you have any idea

 

[haruspex]

But my question is I cannot find the max hieght without angle

As I pointed out, the angle of launch does not give the relationship between horizontal and vertical displacements. The shape in your diagram is not a straight sided triangle.
To find the initial vertical velocity use the vertical acceleration, the height reached and the vertical velocity when at max height?
First, what is the vertical velocity when at max height?
Next, what standard equation relates those four values?

 

[James1019]

As I pointed out, the angle of launch does not give the relationship between horizontal and vertical displacements. The shape in your diagram is not a straight sided triangle.
To find the initial vertical velocity use the vertical acceleration, the height reached and the vertical velocity when at max height?
First, what is the vertical velocity when at max height?
Next, what standard equation relates those four values?

Can you give me the answer or the equation?
It is a little bit confusing

 

[haruspex]

Can you give me the answer or the equation?
It is a little bit confusing

There is a standard set of "SUVAT" equations for constant acceleration. Five variables, s, u, v, a, t, where u is initial velocity and v is final velocity. There are five equations, each involving four of the variables. I feel sure you have been taught these.

Consider first the part of the trajectory from launch to max height. I asked you to find the initial vertical velocity given vertical acceleration, vertical displacement (height reached) and final vertical velocity (vertical velocity at maximum height).
Do you know a SUVAT equation involving those four?
What is the vertical velocity at maximum height?

 

[Hak]

Can you give me the answer or the equation?
It is a little bit confusing

I am only a student and not an expert, but projectile motion is the subject I master the most and am most familiar with.
As you know, this type of motion describes a parabola, with a composition of two motions: rectilinear along the horizontal axis, uniformly accelerated rectilinear (with acceleration equal to $g$) along the vertical axis.
One way to solve this problem is to find the analytical expression of the equation of the maximum height $y_{\text{max}}$ as a function of the initial velocity (and, in particular, of the latter's vertical component). As @haruspex pointed out well, what happens to the velocity of the projectile when the body reaches its maximum height, i.e. the vertex of the parabola? Once you have succeeded in deduce it, you can calculate (as a function of the initial vertical velocity) the time it takes for the projectile to reach the maximum height. Substitute this time into the equation of motion of the body along the vertical axis (analogous to that of a body thrown upwards with vertical velocity and subject only to the downward acceleration of gravity, i.e. uniformly decelerated motion) and you will find the expression for the maximum altitude. From there, you can calculate the initial vertical velocity.
In addition, you are given the range, i.e. the maximum horizontal distance, from the text. Carry out a procedure similar to the previous one and you will obtain an expression as a function of the horizontal velocity (constant) and the initial vertical velocity. You have already found this velocity previously, replace it, and thus obtain the horizontal velocity.
There is no need for angles or trigonometric functions of any kind.

I agree with @haruspex, it is much better to work in symbols than with numbers initially. If you can, substitute numbers only at the end.

 

[Hak]

There is a standard set of "SUVAT" equations for constant acceleration. Five variables, s, u, v, a, t, where u is initial velocity and v is final velocity. There are five equations, each involving four of the variables. I feel sure you have been taught these.

Consider first the part of the trajectory from launch to max height. I asked you to find the initial vertical velocity given vertical acceleration, vertical displacement (height reached) and final vertical velocity (vertical velocity at maximum height).
Do you know a SUVAT equation involving those four?
What is the vertical velocity at maximum height?

Sorry, I had not seen this message. Sorry if I took the liberty to comment, and I also apologise if it would not be possible for me to comment, as I am only a student.

 

[haruspex]

Sorry, I had not seen this message. Sorry if I took the liberty to comment, and I also apologise if it would not be possible for me to comment, as I am only a student.

Your post is fine, though my preference with someone who is struggling is just to deal with the next step or two in each post. Otherwise they can get overwhelmed or try steps out of sequence.

 

[Hak]

Your post is fine, though my preference with someone who is struggling is just to deal with the next step or two in each post. Otherwise they can get overwhelmed or try steps out of sequence.

Yes, right, you are absolutely right. However, I am Italian and I was never taught the SUVAT method. It often happens that we cannot remember all the final equations by heart, so it happens just as often that we have to derive the equations needed for the unfolding from the time laws of motion.

 

[Lnewqban]

Can you give me the answer or the equation?
It is a little bit confusing

Welcome, @James1019 !

It will not be confusing if you take a closer look to nature.
Jumping frogs don't know anything about angles.

Carefully observe the pure vertical velocity of the midbody of the frog in this animation.
Then, try to answer the good questions of post #7 above.