Question

Two bodies are thrown at angle $\theta$ and $(90-\theta )$ from the same point with same velocity $25m{s}^{-1}$. If the difference between their maximum heights is 15m, the respective maximum heights are $(g=10m{s}^{-2})$

• A. $\frac{185}{8}m$ and $\frac{65}{8}m$
• B. $\frac{125}{4}m$ and $\frac{65}{4}m$
• C. 40 m and 25 m
• D. 25 m and 40 m

Answer 1

The correct option is A $\frac{185}{8}m$ and $\frac{65}{8}m$

Let $1^{st}$ body makes angle $=\theta$

$H_1+H_2=\frac{u^2{\sin }^2\theta }{2g}+\frac{v^2{\sin }^2(90-\theta )}{2g}$

$=\frac{u^2}{2g}({\sin }^2\theta +{\cos }^2\theta )$

$\frac{u^2}{2g}$

$H_1+H_2=\frac{25\times 25}{2\times 10}=\frac{125}{4}$m ……….$\left(1\right)$

$H_1-H_2=15$ (given) ………$\left(2\right)$

Add both

$2H_1=\frac{125}{4}+15$

$H_1=\frac{125}{8}+\frac{15}{2}$

$H_1=\frac{185}{8}$

$H_2=\frac{65}{8}$m.