Question

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?

Answer 1

The ball's velocity is 5.88 m/s after 0.6 seconds and 11.76 m/s after 1.2 seconds. It falls a distance of 1.764 m within the first 0.6 seconds, and 7.056 m within the first 1.2 seconds.

Step-by-step explanation:

Falling Steel Ball: Velocity and Distance Calculations

The problem at hand involves a steel ball dropped from rest, which is subject to the acceleration due to gravity (g = 9.8 m/s2). To solve such physics problems, we use kinematic equations that describe the motion of the ball.

(a) Velocity after 0.6 seconds of release

The velocity (v) of a freely falling object released from rest can be found using the equation: v = g × t. Therefore, after 0.6 seconds:

v = 9.8 m/s2 × 0.6 s = 5.88 m/s

(b) Velocity after 1.2 seconds of release

Similarly, after 1.2 seconds:

v = 9.8 m/s2 × 1.2 s = 11.76 m/s

(c) Distance fallen in the first 0.6 seconds

To calculate the distance (d) fallen, we use the equation: d = 0.5 × g × t2. So, in the first 0.6 seconds:

d = 0.5 × 9.8 m/s2 × (0.6 s)2 = 1.764 m

(d) Distance fallen in the first 1.2 seconds

In the first 1.2 seconds:

d = 0.5 × 9.8 m/s2 × (1.2 s)2 = 7.056 m

Answer 2

a) v = 5.88 m/s

b) v = 11.76 m/s

c) s = 1.764 m

d) s = 7.056 m

Step-by-step 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