Question

A 150kg ball, travelling at 30m/s, strikes the palm of a player’s hand and is stopped in 0.05second. Find the force exerted by the ball on the hand​

Answer 1

Answer:

hey mate! here's your answer.

Given - m= 150g , u = 30m/s , t = 0.06s and v = 0m/s

Given - m= 150g , u = 30m/s , t = 0.06s and v = 0m/sForce (F) =150  x 10-3( 0- 30 )/0.06

Given - m= 150g , u = 30m/s , t = 0.06s and v = 0m/sForce (F) =150  x 10-3( 0- 30 )/0.06       F = - 75N

Given - m= 150g , u = 30m/s , t = 0.06s and v = 0m/sForce (F) =150  x 10-3( 0- 30 )/0.06       F = - 75NThe negative sign indicates that the direction of force is opposite to the direction of  motion of the ball.

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Answer 2

✎ Given-✎

☞Mass of Ball = 150kg

☞Initial Speed of the ball = 30m/s

☞Final Speed of ball = 0m/s

☞Time taken= 0.05 sec

   

✎ To FInd-✎

☞Force -?

     

✎ Formulas used-✎

☞ $\boxed{ \sf{F= m\times a }}$

☞ $\boxed{\sf{a= \dfrac{v-u}{t} }}$

where,

F= force,

m= mass,

a= acceleration,

v= final speed,

u= initial speed.

     

✎  SoLution-✎

☞At first we will find acceleration ,

$\bf{a = \dfrac{v-u}{t} }$

$\sf{\implies a = \dfrac{v-30}{0.05} }$

$\sf{\implies a= -600}$

☞Now,

F = m×a

F = 150 × (-600)

F= 90000 N

$\underline{\pink{\bf{\therefore Force \ exerted \ by \ ball \ on\ hand \ is \ 90000N}}}$.

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