Question

Two particles are projected in air with speed $u$ at angles ${\theta }_1$ and ${\theta }_2$ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then which one of the following is correct?
(where $T_1$ and $T_2$ are the time of flight.)

• A. ${\theta }_1$ > ${\theta }_2$
• B. ${\theta }_1$ = ${\theta }_2$
• C. $T_1$ < $T_2$
• D. $T_1$ = $T_2$

Answer 1

Answer: (A)

${\theta }_1$ > ${\theta }_2$

Given that, Two particles are projected in air at speed $u$ at angles ${\theta }_1$ and ${\theta }_2$ to the horizontal.

Also given that height reached by the first particle is greater than that of the second

$⟹H_1>H_2$

Where, $H_1$ is the height reached by first particle and $H_2$ is the height reached by second particle.

$H_1=\frac{u^{2}si{n}^2{\theta }_1}{2g}$

$H_2=\frac{u^{2}si{n}^2{\theta }_2}{2g}$

We have $\frac{u^{2}si{n}^2{\theta }_1}{2g}>\frac{u^{2}si{n}^2{\theta }_2}{2g}$

$⟹si{n}^2{\theta }_1>si{n}^2{\theta }_2$

$⟹(sin{\theta }_1-sin{\theta }_2)(sin{\theta }_1+sin{\theta }_2)>0$

$⟹sin{\theta }_1>sin{\theta }_2$

$⟹{\theta }_1>{\theta }_2$

Let $T_1$ and $T_2$ be the time of flight of first and second particle

$T_1=\frac{2usin{\theta }_1}{g}$

$T_2=\frac{2usin{\theta }_2}{g}$

if $sin{\theta }_1>sin{\theta }_2$

then, $T_1>T_2$

Hence, option $\left(A\right)$ is correct.