Question

Two particles are projected from ground with a speed of $40\text{m/s}$ at complimentary angles. If the difference in the maximum height of the particles is $50\text{m}$, then the height of the individual particles is

• A. $H_1=10\text{m};H_2=75\text{m}$
• B. $H_1=65\text{m};H_2=15\text{m}$
• C. $H_1=35\text{m};H_2=85\text{m}$
• D. $H_1=25\text{m};H_2=75\text{m}$

Answer 1

The correct option is B $H_1=65\text{m};H_2=15\text{m}$

We have been given, $H_1-H_2=50m.........\left(1\right)$

The formula for maximum height is given by,
$H=\frac{u^2{\sin }^2\theta }{2g}$
$\therefore H_1=\frac{u^2{\sin }^2\theta }{2g}$
$H_2=\frac{u^2{\cos }^2\theta }{2g}$

Take, $H_1+H_2=\frac{u^2{\sin }^2\theta }{2g}+\frac{u^2{\cos }^2\theta }{2g}$
$H_1+H_2=\frac{u^2}{2g}=\frac{40\times 40}{2\times 10}=80m$
$H_1+H_2=80m.........\left(2\right)$

By adding $\left(1\right)$ and $\left(2\right)$, we get
$2H_1=130$
$\Rightarrow H_1=65m$
Similarly, we get $H_2=15m$