Question

Two balls are thrown simultaneously at two different angles with the same speed $v_o$ from the same position so that both have equal ranges. If $H_1$ and $H_2$ be the maximum heights attained in the two cases, then the sum $H_1+H_2$ is equal to

• A. $\frac{{v_o}^2}{2g}$
• B. $\frac{{v_o}^2}{g}$
• C. $\frac{{2v_o}^2}{g}$
• D. $\frac{{v_o}^2}{4g}$

Answer 1

The correct option is A $\frac{{v_o}^2}{2g}$

Let, R be the range of the balls and $\theta$ be the angle of projection. We know range is given by

$R=\frac{v_0^2\sin 2\theta }{g}$ ..........(1)

Here, the range is equal for both the balls, but the angles of projection are different, we know

$\sin 2\theta =sin(180-2\theta )$

$\sin 2\theta =sin2(90-\theta )$

Therefore, the angles of projection for two balls must be, $\theta$ and $(90-\theta )$.

Now, the maximum height attained by first ball will be,

$H_1=\frac{v_0^2{\sin }^2\theta }{2g}$ ................(2)

and the maximum height attained by second ball will be,

$H_2=\frac{v_0^2{\sin }^2(90-\theta )}{2g}$ ...............(3)

adding equations (2) and (3) we get,

$H_1+H_2=\frac{v_0^2{\sin }^2\theta }{2g}+\frac{v_0^2{\sin }^2(90-\theta )}{2g}$

$H_1+H_2=\frac{v_0^2{\sin }^2\theta }{2g}+\frac{v_0^2{\cos }^2\theta }{2g}$

$H_1+H_2=\frac{v_0^2}{2g}(si{n}^2\theta +co{s}^2\theta )$

$H_1+H_2=\frac{v_0^2}{2g}$