Question

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

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

Answer 1

The correct option is A 20 m, 70 m

u = 40 m/s

$H_2=H_1+50$ ...(1)

Here $H_1$ = Height reached by first ball

$H_2=$ Maximum height reached by second ball

$g=9.8m{s}^{-2}$

$H_1=\frac{4^{2}si{n}^2\theta }{2g}$ $H_2=\frac{4^{2}si{n}^2(90-\theta )}{2g}$

$H_1=\frac{(40{)}^{2}si{n}^2\theta }{19.6}$ $H_2=\frac{\left(40{)}^{2}si{n}^2\right(90-\theta )}{19.6}$

Putting value of (2nd) in (1st)

$H_1=\frac{(40{)}^{2}co{s}^2\theta }{19.6}$ = $H_1=\frac{(40{)}^{2}si{n}^2\theta }{19.6}+50$

by solving

$H_1=20m$

$H_2=70m$