Question

A wicket keeper catches a ball moving at 30 m/s. if he moves his hand as the ball is caught so that it comes to rest over 10 cm what is its acceleration ​

Answer 1

$\\ \\ \large\underline{ \underline{ \sf{ \red{given:} }}}$

• Ball is initially moving at 30m/sec.

• Ball is caught by the wicket Keeper and it comes to rest over 10cm.

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

• Acceleration.

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

➛ Initial velocity (u) = 30m/sec

➛ Final velocity (v) = 0m/sec

➛ Displacement (s) = 10cm = 0.1m

➛ Acceleration (a) = ?

➢ By 3rd Kinematical equation ,

$\\ \boxed{ \bf {v}^{2} = {u}^{2} + 2as }$

Putting values , we get

$\\ \sf \: {0}^{2} = {30}^{2} + 2a(0.1) \\ \\ \sf \: 0 = 900 + 0.2a \\ \\ \sf \: 0.2a = - 900 \\ \\ \sf \: a = \frac{ - 900}{0.2} \\ \\ \boxed{ \sf \: \green{ a = - 4500m {sec}^{ - 2} }}$

Hence , acceleration is -4500m/sec².

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NOTE :-

• Here negative sign indicates that acceleration is in opposite direction to the motion of ball.

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MORE TO KNOW :-

➢ 1st Kinematical equation ,

• v = u + at

➢ 2nd Kinematical equation ,

• s = ut+(1/2)at²

Answer 2

Given data:

• Distance (s) = 10 cm = 0.1 m
• Initial velocity (u) = 30 m/s
• Final velocity (v) = 0

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To find:

Acceleration of the ball.

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Solution:

Let the time be x.

We know that,

2as = v² – u²

⟹ 2(0.1)a = 0 – (30)²

⟹ 0.2a = –900

⟹ a = –900/0.2

= –4500 m/s²

Therefore –4500 m/s² is the required solution.

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