Question

Caleb is at the top of a tower which is 21 m high. He throws a ball horizontally away from the tower. If the ball lands on the ground 15 m away from the tower: (a) How long does it take the ball to reach the ground? (b) What is the velocity of the ball as it leaves Caleb’s hand? (c) What is the final velocity of the ball the instant before it hits the ground?

Answer 1

Answer:

Given below:

Explanation:

x = utcos∝ , y = utsin∝ - gt²/2 , the ball is thrown horizontally , ∝ = 0

x = ut  = 15 , y =  - gt²/2 = -21

a) gt²/2 = 21 , t = √42/g = √4.2 sec

b) u = 15/√4.2  m/s

vx= u = 15/√4.2 , vy = gt = 10√4.2

c) v² = 15²/4.2 + 420 = 473.5 , v = √473.5 = 21.76 m/s

Hope it helps.