Question

Range of a projectile projected at angle θ and (90° - θ) is same. If in both situations the time of flight is t₁ and t₂, then show that the range will directly proportional to product of t₁ and t₂.

Answer 1

\(\because\ \text{Horizontal}\ \text{range},\ R = \frac{u^2\sin\ 2\theta}{g}\ .......(1)\)

Time of flight for projection angle \(\theta\),

\(t_1 = \frac{2u \sin\theta}{g}\ ....(2)\)

and that for \((90^\circ - \theta),\) 

\(t_2 = \frac{2u\sin(90^\circ - \theta)}{g} = \frac{2u\cos\theta}{g}\ .....(3)\) 

On multiplying equation (2) and (3)

\(t_1t_2 = \frac{2u\sin\theta}{g}\times\frac{2u\cos\theta}{g}\) 

or \(t_1t_2 = \frac{2u^2.\sin\ 2\theta}{g^2}\) 

or \(t_1t_2 = \frac{2R}{g}\)

or \(R = \frac{g}{2}.t_1t_2\)

\(\therefore \ R \propto t_1t_2\)