Aball is thrown from the top of a building with an initial velocity of 21.9 m/s straight upward, at an initial height of 51.6 m above the ground. the ball just misses the edge of the roof on its way down, as shown in the figure. (a) determine the time needed for the ball to reach its maximum height. (b) determine the maximum height. (c) determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant. (d) determine the time needed for the ball to reach the ground. (e) determine the velocity and position of the ball at t = 5.35 s.

Part A)

when ball will reach to highest point then it's speed will become zero

so we can use kinematics to find the time

Part b)

for finding the maximum height we can use another kinematics equation

so it will rise to 1.12 m from the point of projection

Part C)

Ball will take double the time which it take to reach the top point.

So here the time to reach the top is 2.23 s

so time taken by the ball to reach at same point after projection is given as

Since ball have reached to same point so the final velocity must be same as initial velocity

so we have

downwards

Part d)

when ball reached to the bottom

displacement of ball = -51.6 m

now by kinematics we have

by solving above equation we have

now for the velocity at that instant we have

so its velocity is 38.6 m/s downwards

Part e)

for the position of ball at t = 5.35 s we can use

so it is 23.1 m below the initial position from which it is thrown

now for the velocity we can say

so it will be moving downwards with speed 30.53 m/s

true tan speed = ang vel x radius