Question

a player stops a cricket ball of mass 0.170kg in 1/4 by exerting a froce 16.65N with what velocity was the ball moving​

Answer 1

As the player stopped the ball which means

v=0

Mass (m) = 0.170kg

Force (F) =16.65N

Time (t) = 1/4 sec

u = To find

F = ma

F=m×(v-u/t)

F/m × t = u

$\frac{16.65}{0.170} \times \frac{1}{4} \\ = \frac{97.94}{4} \\ = 24.48$

$\huge\fcolorbox{red}{aqua}{u = 24.48}$

I hope it helps ✌.

Answer 2

Answer

The Ball was moving with a velocity of 24.475 m/s

Explanation

Given:-

• Mass of ball, m = 0.170 kg
• Force applied to stop the ball, F = -16.65 N
• time taken to stop the ball, t = 1/4 sec
• final velocity of ball, v = 0  m/s

To find:-

• Initial velocity of ball, u =?

Formula required:-

• Expression of Newton's second law of motion

        F = ma

• First equation of motion

       v = u + at

[ Where F is force, m is mass, a is acceleration, v is final velocity, u is initial velocity and t is time ]

Solution:-

Using newton's second law expression

→ F = ma

→ -16.65 = 0.17 × a

→ a = -97.9  m/s²

Using the first equation of motion

→ v = u + at

→ 0 = u + -97.9 × 1/4

→ u = 24.475 m/s

Therefore,

• Ball was moving with a velocity of 24.475 m/s.