A ball is thrown from the top of a building with an initial velocity of 22.0 m/s

Question

A ball is thrown from the top of a building with an initial velocity of 22.0 m/s straight upward, at an initial height of 55.2 m above the ground.

(a) Determine the time needed for the ball to reach its maximum height. ___s

(b) Determine the maximum height. _____m

(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. Time ____s, Velocity _____ m/s

(d) Determine the time needed for the ball to reach the ground._______s

(e) Determine the velocity and position of the ball at t = 5.39 s. Velocity_____m/s, Position___m

Explanation / Answer

Given

Initial velocity u = 22 m/s

Angle of projection = 90o

Height of the building h = 55.2 m

Solution

A)

Time needed to reach this distance t = u/g

= 22/9.8

= 2.24 s

B)

Maximum height reached H = u2/2g

= 222/2x9.8

= 24.69 m

The maximum height from ground = 24.69 + 55.2 = 79.89 m

C)
Time of flight to reach the building tf = 2t = 2u/g = 4.48 s

V = 0 + gt

= 0 – 9.8 * 2.24

= 22 m/s

D) total distance downward to earth = 79.89 m

S = u’t + 1/2gt’2

79.89 = 0 + 9.8t’2/2

t’ = 4.04 s

total time taken T = t + t’ = 2.24 + 4.04 = 6.28 s

E)

During the time 5.39s the first 2.24 are spent in reaching the maximum height with acceleration a =-g after that it moves with acceleration a = +g

so at t = 5.39 – 2.24 = 3.15 seconds

the velocity v = u’ + gt

= 0 + 9.8*3.15

= 30.87 m/s

the position s = u’t + ½ gt2

s =0 + ½ gx 3.152

= 48.62 m

Which means the ball is at a height h = 79.89 – 48.62 = 31.27 m

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