A ball of mass 50 ,g is dropped from a height of 20 ,m . A boy on ground hits ball vertically upward

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4-1.Newton's Laws of Motion

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A ball of mass $50 \,g$ is dropped from a height of $20 \,m$. A boy on the ground hits the ball vertically upwards with a bat with an average force of $200 \,N$, so that it attains a vertical height of $45 \,m$. The time for which the ball remains in contact with the bat is ........... of a second [Take $g=10 \,m / s ^2$ ]

A

$1 / 20^{\text {th }}$

B

$1 / 40^{\text {th }}$

C

$1 / 80^{\text {th }}$

D

$1 / 120^{\text {th }}$

Solution

(c)

$\text { Using } v^2 =u^2+2 as$

$v_1 =\sqrt{2 g(20)}=20 \,m / s$

$v_2 =\sqrt{2 g(45)}=30 \,m / s$

$\text { Impulse } =F \Delta t=m\left(\vec{v}_2-\overrightarrow{v_1}\right)$

$\Rightarrow 200 t =\frac{50}{1000}(20-(-30) v_1 \downarrow$

$t=\frac{5}{400}=\frac{1}{80} \,s$

Standard 11

Physics

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