# Projectile motion

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1. Oct 12, 2015

### Physics2341313

I've put the problem statement below and worked it out. I typically don't post questions like this as they're a lot to go through, but I am wondering if I have worked the problem correctly as my book does not have the solution and I feel like I am not understand the material correctly.

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

A shell is fired upward from the top of a building at an angle of $\pi /3$ with the horizontal. Its initial velocity is 60 m/s, and assume it is fired from the point (0, 500) when $t=0$ and $x''(t) = 0$ and $y''(t)=-9.8 m/s^2$ while $0 \le t \le t_2$ is the time when the shell hits the ground.

3. The attempt at a solution

Given that $x''(t) = 0$ and $y''(t) = -9.8$ we have r''(t) = <0, -9.8> integrating the vector-valued function we have $$r'(t) = <C, -9.8t + C> \rightarrow r'(t) = <60cos(\pi/3), -9.8t + 60sin(\pi/3)>$$ using the initial velocity given. Integrating again to get the position we have $$r(t) = <60cos(\pi/3)t + C, -4.9t^2 + 60sin(\pi/3)t + C>$$ $$r(t) = <60cos(\pi/3)t, -4.9t^2 + 60sin(\pi/3)t + 500>$$ using the point (0, 500) given.

Now, are the parametric equations $x(t)$ and $y(t)$ for $r(t)=<x(t),y(t)>$ just $x(t) = 60cos(\pi/3)t$ and $y(t) = -4.9t^2 + 60sin(\pi/3) + 500$?

Solving for the time the shell reaches the maximum height will be the time at which $y'(t) = 0$? so we have: $$-9.8t + 60sin(\pi/3) = 0 \Rightarrow t = 60sin(\pi/3)/9.8 = 5.3 s$$
The maximum height will then be $$y(5.3) = -4.9(5.3)^2 + 60sin(\pi/3)(5.3) + 500 = 637.75 m$$
The time the shell hits the ground is when $y(t) = 0$? so we have: $$0 = -4.9t^2 + 60sin(\pi/3)t + 500 \Rightarrow t = 10.11$$
Horizontal distance: $x(10.11) = 60cos(\pi/3)(10.11) = 303.3 m$

The speed of shell at impact:

$||r'(10.11)|| = \sqrt{(60cos(\pi/3))^2 + (-9.8(10.11) + 60sin(\pi/3))^2} = 55.86 m/s$

Is this the correct way to calculate the speed? As speed is just the magnitude of the velocity?

Last edited: Oct 12, 2015
2. Oct 12, 2015

### PWiz

The angles are different. Am I missing something?

3. Oct 12, 2015

### Physics2341313

No, sorry I mistyped I'll fix it.