Question

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

Answer 1

Answer:

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