Question

A cricket player throws ball in air at 60degree with horizontal axis with K. E=50j .The k.E of ball at heighest point of its flight will be​

Answer 1

Initial kinetic energy of the ball,

• $\sf{K=\dfrac{1}{2}\,mu^2}$

At the highest point the ball will have horizontal velocity only.

• $\sf{v=u\cos\theta}$

Hence kinetic energy of ball at highest point will be,

$\sf{\longrightarrow K'=\dfrac{1}{2}\,mv^2}$

$\sf{\longrightarrow K'=\dfrac{1}{2}\,m(u\cos\theta)^2}$

$\sf{\longrightarrow K'=\dfrac{1}{2}\,mu^2\cos^2\theta}$

$\sf{\longrightarrow K'=K\cos^2\theta}$

Here,

• $\sf{\theta=60^o}$

• $\sf{K=50\ J}$

Then,

$\sf{\longrightarrow K'=50\cos^260^o}$

$\sf{\longrightarrow K'=50\times\dfrac{1}{4}}$

$\sf{\longrightarrow\underline{\underline{K'=12. 5\ J}}}$