# Projectile problem find angles

• mac227
In summary: OK, you're right. It's v=dx/dt, so x=vt. I am having a brain cramp tonight. Sorry.OK, so when you solve for t, you get a function of theta. I believe that the function is:t(theta)=(91.4m)/(v*cos(theta)). Now, plug this into the equation of motion in the y direction and solve for the angle."In summary, the problem involves a rifle that is 'sighted in' for a 91.4-meter target and has a muzzle speed of 596 m/s. There are two possible angles, θ1 and θ2, between the rifle barrel and the horizontal that will allow the bullet
mac227

## Homework Statement

The drawing shows an exaggerated view of a rifle that has been ‘sighted in' for a 91.4-meter target. If the muzzle speed of the bullet is v0 = 596 m/s, there are the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target. One of these angles is so large that it is never used in target shooting. Give your answers as (a) the smaller angle and (b) the larger angle.
(Hint: The following trigonometric identity may be useful: 2 sinθ cosθ = sin 2θ.)

## The Attempt at a Solution

I have no idea how to begin this problem!

I know that vox=vcosθ and voy=vsinθ but do not have sufficient info to begin?

x= 91.4
v=596
ax=0
vx=vox
ay=-9.8 (when going up where voy=? and vy= 0) ay = 9.8 (when voy=0 and vy=?)

I think i need to start off by calculating the time it takes for bullet to reach peak?

You need to use the fact that you know the displacement at the moment the bullet hits the target. If you can get formulas for displacement versus time, you can use that to find the angle.

mac227 said:

## Homework Statement

The drawing shows an exaggerated view of a rifle that has been ‘sighted in' for a 91.4-meter target. If the muzzle speed of the bullet is v0 = 596 m/s, there are the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target. One of these angles is so large that it is never used in target shooting. Give your answers as (a) the smaller angle and (b) the larger angle.
(Hint: The following trigonometric identity may be useful: 2 sinθ cosθ = sin 2θ.)

## The Attempt at a Solution

I have no idea how to begin this problem!

I know that vox=vcosθ and voy=vsinθ but do not have sufficient info to begin?

x= 91.4
v=596
ax=0
vx=vox
ay=-9.8 (when going up where voy=? and vy= 0) ay = 9.8 (when voy=0 and vy=?)

I think i need to start off by calculating the time it takes for bullet to reach peak?
OK. You're right that you need to start off calculating the time it takes to reach peak. Try this: Can you calculate the time it takes to travel all 91.4m in the x direction? If so, can you find the time at the peak? If so, you can plug your result for t_peak into the equation for the y direction motion and solve for the angles.

HINT: You are not going to be able to get a number for t_peak. You will get a function of theta.

"OK. You're right that you need to start off calculating the time it takes to reach peak. Try this: Can you calculate the time it takes to travel all 91.4m in the x direction? If so, can you find the time at the peak? If so, you can plug your result for t_peak into the equation for the y direction motion and solve for the angles.

HINT: You are not going to be able to get a number for t_peak. You will get a function of theta."

Well, to calculate the time it takes to travel 91.4m wouln't i need the x velocity, which i cannot find without the angle (vox=v(cosθ)) I do not believe i can use x=v/t because the given velocity is the magnitude of velocity. I know of no other formulas I can use here, nor have I been able to find any.

Would x=vo(t)+.5(a)t2 work here??

so, 91.4=596(t) t=0.153s

If t=.153 then the time it takes to reach peak would be .076

not sure if this is correct thus far, but if it is I am stuck here

mac227 said:
"OK. You're right that you need to start off calculating the time it takes to reach peak. Try this: Can you calculate the time it takes to travel all 91.4m in the x direction? If so, can you find the time at the peak? If so, you can plug your result for t_peak into the equation for the y direction motion and solve for the angles.

HINT: You are not going to be able to get a number for t_peak. You will get a function of theta."Well, to calculate the time it takes to travel 91.4m wouln't i need the x velocity, which i cannot find without the angle (vox=v(cosθ)) I do not believe i can use x=v/t because the given velocity is the magnitude of velocity. I know of no other formulas I can use here, nor have I been able to find any.

This is why I said you won't get a number when you solve for t, but instead an equation in terms of theta that you then plug into the kinematic equations for the y direction motion.

Would x=vo(t)+.5(a)t2 work here??

so, 91.4=596(t) t=0.153s

If t=.153 then the time it takes to reach peak would be .076

not sure if this is correct thus far, but if it is I am stuck here
EDIT: Had an error here. See my next post for correct advice.

Last edited:
G01 said:
This is why I said you won't get a number when you solve for t, but instead an equation in terms of theta that you then plug into the kinematic equations for the y direction motion.

a in the x direction is 0, so that answer is wrong, but you can use that equation. Use it and solve for t in terms of theta. What do you get?

x=vo(t)+.5(a)t2
so, 91.4=596(t) + .5(0)t2
which becomes 91.4=596(t)

t=0.153s

not sure what you mean by solve for t in terms of theta?

mac227 said:
x=vo(t)+.5(a)t2
so, 91.4=596(t) + .5(0)t2
which becomes 91.4=596(t)

t=0.153s

not sure what you mean by solve for t in terms of theta?

Sorry. I had an error in my above post. I was in a rush. Apologies.

The reason that answer is wrong is because you are using $v_o$ when you should be using $v_o_x=v_o cos\theta$.

You cannot solve for t and get a number, but that doesn't mean you can't solve for t and get an equation in terms of theta.

Let's do that. Solve for t as a function of theta. (Get t on one side of the equation and theta on the other.) What do you get?

ok so then x=vo(cosθ)(t)+.5(a)t2

cosθ=.5(a)t/x θ=cos-1(.5at/x) ?

not sure about what to do with cos

mac227 said:
ok so then x=vo(cosθ)(t)+.5(a)t2

cosθ=.5(a)t/x θ=cos-1(.5at/x) ?

not sure about what to do with cos

No. a in the x-direction is zero. So you have:

$$x=(v_o cos\theta) t$$

Now, get t by itself on one side.

By the way, here is what I'm trying to walk you through:

You have two unknowns, t and theta, thus you are going to need two equations before you can solve for either one.

1. First Equation: I want you to find an equation relating t and theta from the information about the x-direction motion. This is what we are doing now.

2. Second equation. You are going to do the same thing for the y-direction.

3. Then we will solve both equations simultaneously to find theta (and t, but they don't ask for that).

Sorry, I spoke too soon. This question is quite easy, although not all questions like this are. I rashly concluded that the question was not well thought out, but I had not understood it.

Although now I can't figure out how to get the second angle.

Okay, got my brain in order. I had a formula with arccos originally that confused me. Here are some tips:

Get two linear equations in t, one with v_0 cos(theta), one with v_0 sin(theta). Then eliminate t and solve for theta.

Last edited:
first equation: $$x/(v_o cos\theta)= t$$

Second equation: y=voysinθt+.5(-9.8)t2
y=voysinθt + -4.9t2 cancel out the t's ?
( y-voy sinθ ) / -4.9 = t ...??

mac227 said:
first equation: $$x/(v_o cos\theta)= t$$

Second equation: y=voysinθt+.5(-9.8)t2
y=voysinθt + -4.9t2 cancel out the t's ?
( y-voy sinθ ) / -4.9 = t ...??

OK. Good. That equation is true, but we don't know y, so it introduces another unknown into the problem, which we definitely don't want to do if we don't have to.

How about another kinematic equation, one not involving y?

So an equation without the y distance or y units altogether?

vy=voy+(ay)t

(vy-voy)/ay= t

Just and without the y distance variable. I can still have the y velocity involved.

mac227 said:
So an equation without the y distance or y units altogether?

vy=voy+(ay)t

(vy-voy)/ay= t

Great!

Now we know the y-velocity of the ball at t=0 (v_oy)

Do we know the y-velocity of the ball at a second point?

we know that the velocity of the ball is also 0 at its peak and when it comes to rest...

So a=-9.8 v=0 and v0y=? when calculating the ball's rise to its peak.

On its way down voy=0 and vy=? a=9.8

I have trouble seeing how to use (vy-voy)/ay= t too many unknowns?

mac227 said:
we know that the velocity of the ball is also 0 at its peak and when it comes to rest...

So a=-9.8 v=0 and v0y=? when calculating the ball's rise to its peak.

On its way down voy=0 and vy=? a=9.8

I have trouble seeing how to use (vy-voy)/ay= t too many unknowns?

$$v_{oy}=v_o sin\theta$$

Remember?

Plug that in for v_oy and we have two unknowns, t and theta. Luckily we have two equations! You can then eliminate t from the equations and solve for theta.

i plugged $$v_{oy}=v_o sin\theta$$ and (vy-voy)/ay= t into vy=voy+(ay)t
to get:

vy=voysin(theta)+ay(vy-voy/ay)

sin`-1 * ((vy-ay(vy-voy/ay))/voy)=theta

The problem involves x displacement,y displacement, velocity and angle of projection.
Here y displacement is zero.
Using v = vsinθ - gt and x = vcosθ*t, you can write a combined equation as
y = x*tanθ - 1/2*g*x^2/v^2cos^2(θ) or
y = x*tanθ - 1/2*g*x^2v^2*sec^2(θ).
Put sec^2(θ) = 1 + tan^2(θ). Solve the quadratic to find tanθ and proceed

To eliminate t, we need two equations where t represents the same value. This will only happen when we are dealing with one point.

If the y-displacement at the target is not zero, we can't assume that time to peak is half the time to target. We can assume this if the target y-displacement is zero.

Using the velocity formula for the time to peak, we can't eliminate t using a formula where t is the time to target.

I think we must take the y-displacement to be zero. Then we have these two equations:

$$v_0 cos(\theta) t_{target} = 91.4,$$

$$v_0 sin(\theta) t_{target} - (1/2) 9.8 t_{target}^2 = 0.$$

Mac227 did get these formulas but did not set y to 0. Doing so will lead to a solution. If y_target is not zero, one gets into all manner of trouble.

rl.bhat said:
The problem involves x displacement,y displacement, velocity and angle of projection.
Here y displacement is zero.
Using v = vsinθ - gt and x = vcosθ*t, you can write a combined equation as
y = x*tanθ - 1/2*g*x^2/v^2cos^2(θ) or
y = x*tanθ - 1/2*g*x^2v^2*sec^2(θ).
Put sec^2(θ) = 1 + tan^2(θ). Solve the quadratic to find tanθ and proceed

you lost me here

sec^2(θ) ? does this mean sec(θ) 2 ?

mac227 said:
you lost me here

sec^2(θ) ? does this mean sec(θ) 2 ?

Yes, that is what he means.

## 1. What is a projectile problem?

A projectile problem involves finding the initial velocity, angle, and height of a projectile in motion, usually in a two-dimensional space. It is commonly used in physics and engineering to solve real-world problems such as predicting the trajectory of a ball or a rocket.

## 2. How do you find the angle in a projectile problem?

To find the angle in a projectile problem, you can use the formula: θ = tan^-1 (vy/vx), where θ is the angle, vy is the vertical velocity, and vx is the horizontal velocity. This can be derived from the trigonometric ratios of a right triangle.

## 3. What information is needed to solve a projectile problem?

To solve a projectile problem, you will need the initial velocity of the projectile, the angle at which it is launched, and the height at which it is launched. Additional information such as the acceleration due to gravity and air resistance may also be required depending on the specific problem.

## 4. How do you calculate the time of flight in a projectile problem?

The time of flight in a projectile problem can be calculated using the formula: t = 2vy/g, where t is the time of flight, vy is the vertical velocity, and g is the acceleration due to gravity. This formula assumes that the initial and final height of the projectile are the same.

## 5. Can the same angle produce different ranges in a projectile problem?

Yes, the same angle can produce different ranges in a projectile problem. This is because the range of a projectile is affected by various factors such as the initial velocity, height, and air resistance. Therefore, two projectiles launched at the same angle but with different initial velocities or heights will have different ranges.

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