Question

A cricket ball is hit with velocity $25m{s}^{-1}$, ${60}^o$ above the horizontal. How far above the ground, ball passes over a field $50m$ from the bat(consider the ball is struck very close to the ground)? (Take $\sqrt{3}=1.7$ and $g=10m{s}^{-1}$)

• A. $6.6m$
• B. $9.0m$
• C. $11.6m$
• D. $12.7m$

Answer 1

The correct option is A $6.6m$

equation of trajectory is $y=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{u^2{(\cos \theta )}^2}$

substituting respective values will give us
$y=50\tan 60-\frac{1}{2}\frac{10\times {50}^2}{{25}^2{(\cos 60)}^2}$

$⟹y=5m$