Question

There are two angles of projection for which the horizontal range is the same. Show that the sum of the maximum heights for these two angles is independent of the angles of projection.

Answer 1

Let the two angles be ${\theta }_1$ and ${\theta }_2$ and the projectile velocity be $u$

Horizontal range $R=\frac{u^{2}sin\left(2{\theta }_1\right)}{g}=\frac{u^{2}sin\left(2{\theta }_2\right)}{g}=\frac{u^{2}sin(180-2{\theta }_2)}{g}$

$⟹$ $2{\theta }_1={180}^o-2{\theta }_2$

$⟹$ ${\theta }_2=90-{\theta }_1$

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

Similarly $H_2=\frac{u^{2}si{n}^2{\theta }_2}{2g}=\frac{u^{2}si{n}^2(90-{\theta }_1)}{2g}=\frac{u^{2}co{s}^2{\theta }_1}{2g}$

$\therefore$ $H=H_1+H_2=\frac{u^2(si{n}^2{\theta }_1+co{s}^2{\theta }_2)}{2g}=\frac{u^2}{2g}$