# Projectile Motion: Finding the Angle

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1. Nov 17, 2014

### Jazz

1. The problem statement, all variables and given/known data

A toy gun uses a spring with a force constant of $\small{300 N/m}$ to propel a $\small{10.0\!-\!g}$ steel ball. If the spring is compressed $\small{7.00\ cm}$ and friction is negligible:

At what angles above the horizontal may a child aim to hit a target $\small{3.00\ m}$ away at the same height as the gun?

Data:

$m = 10.0 \times 10^{-3}\ kg$

$k = 300\ N/m$

$x = 7.00 \times 10^{-2}\ m$

$Range = 3.00\ m$

2. Relevant equations

$PE_{spring} = \frac{1}{2}kx^2$

$KE = \frac{1}{2}mv^2$

$v^2 = v_0^2 - 2gh$

$\bar{v}t = d$

3. The attempt at a solution

It's assumed conservation of energy in order to find the initial velocity:

$\Delta PE_{spring} = \Delta KE$

$\sqrt{\frac{kx^2}{m}} = v$

$v= 12.1\ m/s$

Then I broke the velocity and the displacement into their components (I skipped the units for simplicity). I used just half of the range, hence I took $\small{d = 1.50\ m}$:

$\sin(\theta) = \frac{y}{x} \Rightarrow x\sin(\theta) = y$

$\cos(\theta) = \frac{1.50}{x} \Rightarrow x = \frac{1.50}{\cos(\theta)}$

$y = 1.50\tan(\theta) \hspace{35pt}(1)$

I wanted to know the height (y) the ball reached:

$v^2 = v_0^2 - 2gy$

$y = \frac{(12.12\sin(\theta))^2}{(2)(9.8)} = 7.50\sin^2(\theta) \hspace{35pt}(2)$

Combining equations $(1)$ and $(2)$:

$1.50\tan(\theta) = 7.50\sin^2(\theta)$

$\cos(\theta)\sin(\theta) = \frac{1}{5}$

$2\cos(\theta)\sin(\theta) = \frac{2}{5}$

$\sin(2\theta) = \frac{2}{5}$

$2\theta = \arcsin(\frac{2}{5})$

$\theta = 11.79º\ or\ 78.21º$

Then by finding the time I wanted to comprobe whether the results I got are corrects. The time it took the ball to reach its maximum height, its horizontal displacement should have been $\small{d = 1.50\ m}$:

$t= \frac{v\sin(\theta)}{g}= \frac{(12.12\ m/s)\sin(11.79º)}{9.8\ m/s^2} = 0.253\ s$

$v\cos(\theta)t = d$

$(12.12\ m/s)\cos(11.79º)(0.2527\ s) = 3.00\ m$

I don't know why I get this horizontal displacement. I expected to get $\small{d = 1.50\ m}$ and not the whole range.

Thanks!!

Last edited: Nov 17, 2014
2. Nov 17, 2014

### Jazz

I've just discovered that the formula $\small{range = \frac{v^2\sin(2\theta)}{g}}$ is a more straightforward way to solve this problem :) ; but I still don't know what I'm doing wrong above.

At the beginning, I had mistakenly taken the work done by the spring as $Fx$, which yielded a greater velocity that fits the 5.77-degree angle needed for a 3-m range, but the components of the velocities are overestimated, giving me the same problem as the one above.

3. Nov 17, 2014

### Staff: Mentor

In your triangle on the right you've assumed that the projectile will follow a straight-line path at angle θ This is not so! While the launch angle may be θ, the projectile follows a parabolic trajectory.

Your triangle for breaking the initial velocity into its components is okay.

4. Nov 18, 2014

### lep11

Yep, your triangle on the right is incorrect as gneill said. Your calculation of initial velocity looks okay.
vcos(θ)t=d=3m is for horizontal displacement.What is the equation for vertical displacement when the ball hits the target at the same height as the gun?

5. Nov 18, 2014

### Jazz

Thank you, @gneill and @lep11, for answering.

]: True. The triangle is not taking into account the action of $\small{g}$ as the math part does. So I'm left without triangle there. It makes sense.

I hope not to be confusing something, but the displacement in the vertical direction should be zero after hitting the target at the same height as the gun, shouldn't it?

Your help has helped to see where, I think, I was wrong. If a consider just half of the path, when the vertical velocity becomes zero, then an equation for the vertical displacement (that I'd called $\small{y}$) is:

$y = \bar{v}t$

$y = \frac{v\sin(\theta) + 0}{2}t$

$y = \frac{v\sin(\theta)}{2}t \hspace{35pt}(1)$

Since half of the range, $\small{d = 1.50\ m}$, was traveled in $\small{t}$ at $\small{v\cos(\theta)}$, then:

$v\cos(\theta)t = d$

$t =\frac{d}{v\cos(\theta)} \hspace{35pt}(2)$

$y = \left(\frac{v\sin(\theta)}{2}\right) \left(\frac{d}{v\cos(\theta)}\right)$ <--- Combining equations $(1)$ and $(2)$

$y= \frac{d\tan(\theta)}{2}$

That is $\small{y}$ and not $d\tan(\theta)$. Basically I had mistakenly taken the max. height traveled at constat $\small{v\sin(\theta)}$, which in turn meant having changed just one side of the equation when I was trying to solve the problem.

6. Nov 18, 2014

### lep11

You have left g out of your equations.Which kinematics equation relates acceleration=g, displacement, time, and initial velocity to each other?

Last edited: Nov 18, 2014
7. Nov 18, 2014

### Jazz

I guess it should be $y = v\sin(\theta)t - \frac{1}{2}gt^2$.

8. Nov 18, 2014

### lep11

You are right.

Last edited: Nov 18, 2014
9. Nov 18, 2014

### Jazz

Using a geometric reasoning for finding $y$ wasn't good at all |: .

10. Nov 18, 2014

### lep11

y= y0+vsin(θ)t−0.5gt^2 now y=y0 so you can write 0=vsin(θ)t−0.5gt^2 (where t is the total flight time of the ball)

0=vsin(θ)t−0.5gt^2
vcos(θ)t=d=3m
Can you now solve for θ? You have two equations and two unknowns.

Last edited: Nov 18, 2014
11. Nov 18, 2014

### Jazz

Well, now I can see what you've been trying to show me. It never came to my mind to use the equation for y along the whole trajectory.

This should be:

$0 = v\sin(\theta)t-\frac{1}{2}gt^2$

$\frac{1}{2}gt^2 = v\sin(\theta)t \hspace{25pt}| \hspace{25pt} v\cos(\theta)t = d$

$\frac{1}{2}gt = v\sin(\theta) \hspace{33pt}| \hspace{35pt} t = \frac{d}{v\cos(\theta)}$

$\left(\frac{1}{2}g\right)\left(\frac{d}{v\cos(\theta)}\right) = v\sin(\theta)$

$\frac{gd}{v^2} = 2\cos(\theta)\sin(\theta)$

$\frac{gd}{v^2} = \sin(2\theta)$

$\theta = \frac{\arcsin\left(\frac{gd}{v^2}\right)}{2}$

$\theta = \frac{\arcsin\left(\frac{(9.8\ m/s^2)(3.00\ m)}{(12.12\ m/s)^2}\right)}{2} = 5.77º$

Not that bad (: