# Thread: [SOLVED] Physics - Projectile Motion, Vectors

1. ## [SOLVED] Physics - Projectile Motion, Vectors

A playground is on the flat roof of a city school, 5.1 m above the street below (see figure). The vertical wall of the building is h = 6.50 m high, forming a 1.4-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of θ = 53.0°; above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.

(a) Find the speed at which the ball was launched.

There are actually 3 parts to this problem. Currently I only need help with part (a) because I need that to answer the other 2 parts, but I may need help with other parts as well.

The program I am using to submit my work says my answer differs within 10% of the correct value but I have rechecked my answer several times and found no mistakes and I didn't round off until the final answer. What am I doing wrong?

Here are the calculations I have carried out to find the initial velocity:

I used the equation $\displaystyle y_f = y_i + v_{yi} t + (\frac{1}{2}) a_y t^2$

where $\displaystyle a_y = g = -9.80m/s^2$, the acceleration vector due to gravity; $\displaystyle v_{yi}$ is the is the y component of the initial velocity vector; $\displaystyle y_i$ is the y coordinate (the height) of the initial position of the ball; $\displaystyle t$ is time; and $\displaystyle y_f$ is the y coordinate of the final position of the ball.

Since it takes the ball takes 2.20 s to reach a point vertically above the wall, I substituted $\displaystyle t = 2.20s$ and $\displaystyle y_f = 6.50m$, the height of the wall, into the above equation. $\displaystyle y_{i}$ would of course be 0m since the ball starts at ground level.

$\displaystyle \Rightarrow\ 6.50m = 0m + v_{yi}(2.20s) + \frac{1}{2}(-9.80m/s^2)(2.20s)^2$

$\displaystyle \Rightarrow\ 6.50m + 23.716m = v_{yi}(2.20s)$

$\displaystyle \Rightarrow\ v_{yi} = \frac{30.216m}{2.20s} \approx 13.73m/s$ Note: I did not actually round off here for my next calculation.

$\displaystyle sin(\theta_{i}) = \frac{v_{yi}}{v_i}\ \Leftrightarrow v_i = \frac{v_{yi}}{sin(\theta_i)}$

So $\displaystyle v_i = \frac{13.7345m/s}{sin(53.0^{\circ})} \approx 17.2m/s$

I keep getting this same answer. What am I doing wrong?

2. Here are the other two questions, but I'm pretty sure I need the initial velocity first to answer them though. Please help.

(b) Find the vertical distance by which the ball clears the wall.

(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

3. Never mind, I figured out my mistakes for the initial velocity. The correct answer was 18.1m/s. Problem solved.