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 [Take g= 10 m/s $^{2}$ ]

1/20

^{t h}

of a second

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1/40

^{t h}

of a second

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1/80

^{t h}

of a second

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1/120

^{t h}

of a second

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 1/80 $^{t h}$ of a second
$\begin{matrix} V e l o c i t y o f t h e b a l l j u s t b e f o r e h i t t i n g t h e b a t = V_{1} {V_{1}}^{2} = 0^{2} + 2 g s = 2 \times 10 \times 20 = 400 V_{1} = - 20 m / s, V_{1} i s d o w n w a r d s h e i g h t a t t a i n e d b y t h e b a l l a f t e r b e i n g h i t : - 45 m T h e s p e e d j u s t a f t e r b e i n g h i t b y t h e b a t = V_{2} \Rightarrow 0 = {V_{2}}^{2} - 2 g s \Rightarrow {V_{2}}^{2} = 2 \times 10 \times 45 = 900 V_{2} = 30 m / sec (u p w a r d s) C h a n g e i n v e l o c i t y = V_{2} - V_{1} = 30 - (- 20) = 50 m / s C h a n g e i n m o m e n t u m = m V_{2} - m V_{1} = m Δ V = \frac{50}{1000} \times 50 = 2.5 k g m / s f o r c e = \frac{Δ p}{Δ t} = 200 N Δ t = \frac{2.5}{200} = \frac{1}{80} sec \end{matrix}$
Hence,
option $(C)$ is correct answer.

Suggest Corrections