playground is on the flat roof of a city school 6.00 m above the street below . The vertical wall of the building is h = 7.00 m high , forming a 1-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 = 53.0 above the horizontal at a point d = 24.0 m from the base of the building wall . The ball takes 2.20 second to reach a point vertically above the wall .
1- Find the speed at which the ball was launched ?
2- Find the vertical distance by which the ball clears the wall ?
3- Find the horizontal distance from the wall to the point on the roof where the ball lands ?
I answer number (1) speed = 18.1 m/s and number 2 vertical distance = 1.13 m
I need help for number 3??

Q1. Remember that the horizontal component of the velocity is completely independent of gravity. It stays constant throughout the entire flight. So if the ball traveled a horizontal distance of 24m in 2.2 seconds, what was its horizontal velocity? You know the ratio of the horizontal component to the total velocity is equal to the cosine of the launch angle (likewise, the vertical component is proportional to the sine of the angle), so you can use that relationship to solve for the total velocity (a.k.a. speed).

\frac{v_{horz.}}{v_{total}}=\cos 53

\frac{v_{vert.}}{v_{total}}=\sin 53

Q2. You also know that h=\frac12gt^2+v_0t+h_0, where h represents the height of the ball, g is the acceleration of gravity (-9.81 m/s^2), v_0 is the initial vertical component of the velocity (which you can calculate using the second formula from above), and h_0 is the initial height of the ball (0m since it started on the ground). So just plug in t=2.2s to find h. Now subtract 7m to get the height above the top of the wall.

Q3. Using the same formula as in Q2 above, now plug in h=6m (the height of the roof) and solve for t. You should get two answers: one at some time less than 2.2 seconds representing the time when the ball crossed the 6m mark on its way up and over the wall, and a second one at some time after 2.2 seconds representing the time when the ball landed on the roof on its way back down. Obviously it's the greater of the two which is important for this question. Once you know how long it took the ball to reach the roof, use the horizontal component of the velocity (which you calculated in Q1) to compute the total horizontal distance the ball traveled during that amount of time. Subtract 24m from that, and you've got your answer.