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 ish= 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 ofθ= 53.0°; above the horizontal at a pointd= 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?