Question

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${}^2$]

• A. 1/20${}^{th}$ of a second
• B. 1/40${}^{th}$ of a second
• C. 1/80${}^{th}$ of a second
• D. 1/120${}^{th}$ of a second

Answer 1

The correct option is C 1/80${}^{th}$ of a second

$\begin{array}{c}Velocityoftheballjustbeforehittingthebat=V_1\\ {V_1}^2=0^2+2gs=2\times 10\times 20=400\\ V_1=-20m/s,V_{1}isdownwards\\ heightattainedbytheballafterbeinghit:-45m\\ Thespeedjustafterbeinghitbythebat=V_2\\ \Rightarrow 0={V_2}^2-2gs\\ \Rightarrow {V_2}^2=2\times 10\times 45=900\\ V_2=30m/\sec \left(upwards\right)\\ Changeinvelocity=V_2-V_1=30-\left(-20\right)=50m/s\\ Changeinmomentum=m{V}_2-m{V}_1=m\Delta V=\frac{50}{1000}\times 50=2.5kgm/s\\ force=\frac{\Delta p}{\Delta t}=200N\\ \Delta t=\frac{2.5}{200}=\frac{1}{80}\sec \end{array}$

Hence,

option $\left(C\right)$ is correct answer.