A steel ball is dropped from a diving platform from rest. Given that g = 9.8 m/s?,

(a) What is the velocity of the ball 0.6 seconds after its release?

(b) What is the velocity of the ball 1.2 seconds after its release?

(c) Through what distance does the ball fall in the first 0.6 seconds of its flight?

(d) How far does it fall in the first 1.2 seconds of its flight?

a) v = 5.88 m/s

b) v = 11.76 m/s

c) s = 1.764 m

d) s = 7.056 m

Explanation:

Given,

The initial velocity of the steel ball, u = o

The acceleration due to gravity, g = 9.8 m/s²

The equations of motion to find the final velocity for a body under free fall with an initial velocity, u = 0 is

v = u + at m/s

v = at m/s

a) At time t = 0.6 s

v = 9.8 x 0.6

= 5.88 m/s

The velocity of the ball 0.6 seconds after its release is, v = 5.88 m/s

b) At t = 1.2 s

v = 9.8 x 1.2

= 11.76 m/s

The velocity of the ball 1.2 seconds after its release is, v = 11.76 m/s

The distance traveled by the free falling body is given by the formula

s = ut + 1/2 at² m

s = 1/2 at² ∵ u = 0

c) At, t = 0.6 s

s = 1/2 x 9.8 x 0.6²

= 1.764 m

The distance ball fall in the first 0.6 seconds of its flight is, s = 1.764 m

d) At, t = 1.2 s

s = 1/2 x 9.8 x 1.2²

= 7.056 m

The distance ball fall in the first 1.2 seconds of its flight is, s = 7.056 m