Question

A ball is thrown straight up with an initial speed of 4.9 m/s. It is in the air for 1 second. a) With what speed does it hit your hand (held in the position at which it released the ball)? b) What is the acceleration on the way up? c) What is the acceleration on the way down? d) What is the acceleration at the top? e) How long did it take to get to the top?

Answer 1

a]

From conservation of energy, the speed with which the ball hits the hand will be equal to the speed of the ball during the launch.

therefore, v = 4.9 m/s

b] Acceleration of the way up is decelerating the ball so that it reaches a velocity of v = 0 m/s at the highest point at t=0.5s

therefore, a = - g = - 9.8 m/s²

c] Acceleration while the ball starts to travel downwards is positive as it accelerates the ball.

therefore, a = +g = 9.8m/s²

d] Acceleration at the top has the same magnitude since this acceleration is acceleration due to gravity which remains the same regardless of the position of the ball.

so, a = 9.8 m/s²

e] Time taken to reach the top = T/2

T = total travel time = 1s

therefore, t = 0.5s.