Question

A ball is thrown at angle $\theta$ and another ball is thrown at an angle $(90-\theta )$ with the horizontal direction from the same point with the same speed $40m{s}^{-1}$. The second ball reaches $50m$ higher than the first ball. Find their individual heights.

• A. $20m,70m$
• B. $25m,75m$
• C. $15m,65m$
• D. $10m,60m$

Answer 1

The correct option is C $15m,65m$

$h=\frac{u^2{\sin }^2\theta }{2g}$

$H=\frac{u^2\left(\sin \right(90-\theta ){)}^2}{2g}$

As $H-h=50$

$\frac{u^2}{2g}[{\cos }^2\theta -{\sin }^2\theta ]=50$

$\cos 2\theta =\frac{50\times 2g}{u^2}$

$\cos 2\theta =\frac{50\times 2\times 10}{{40}^2}$

or, $\cos 2\theta =\frac{5}{8}$

$\therefore h=\frac{u^2}{2g}\left(\frac{1-\cos 2\theta }{2}\right)$

$=\frac{{40}^2}{2\times 10}\left(\frac{1-\frac{5}{8}}{2}\right)$

$=15m$

$H=50+h=50+15=65m$