Question

Two projectiles are thrown with different velocities and at different angles so as to cover the same maximum height. Show that the sum of the times taken by each to reach the highest point is equal to the total time taken by either of the projectiles.

Answer 1

If the two projectiles are thrown with velocities $u_1$ and $u_2$ at angles ${\theta }_1$ and ${\theta }_2$ with horizontal, then their maximum height will be,

$\Rightarrow H_1=\frac{u_1^{2}si{n}^2{\theta }_1}{2g}$

$\Rightarrow H_2=\frac{u_2^{2}si{n}^2{\theta }_2}{2g}$

but $\Rightarrow H_1=H_2$

Therefore, $\Rightarrow \frac{u_1^{2}si{n}^2{\theta }_1}{2g}=\frac{u_2^{2}si{n}^2{\theta }_2}{2g}$

$\Rightarrow u_{1}sin{\theta }_1=u_{2}sin{\theta }_2$ $\left(1\right)$

Times of flight for the two projectiles are

$\Rightarrow T_1=\frac{2u_{1}sin{\theta }_1}{g}$

$\Rightarrow T_2=\frac{2u_{2}sin{\theta }_2}{g}$

Making use of equation $\left(1\right)$ we get

$\Rightarrow T_1=T_2=\frac{2u_{1}sin{\theta }_1}{g}=\frac{2u_{2}sin{\theta }_2}{g}$

Times taken to reach the highest point in the two cases will be ,

$\Rightarrow t_1=\frac{u_{1}sin{\theta }_1}{g}$ and

$\Rightarrow t_2=\frac{u_{2}sin{\theta }_2}{g}$

$\Rightarrow t_1+t_2=\frac{u_{1}sin{\theta }_1}{g}+\frac{u_{2}sin{\theta }_2}{g}=\frac{2u_{1}sin{\theta }_1}{g}=\frac{2u_{2}sin{\theta }_2}{g}$ (by using equation $1$)

So, $\Rightarrow t_1+t_2=timeofflightofeitherprojectile$