A projectile can have the same range R for two angles of projection. If t₁ and t₂ be the times of flights in the two cases, then the product of the two time of flights is proportional to (a) R² (b) 1/R² (c) 1/R (d) R

Answer: (d). For angles θ and 90° - θ: t₁ t₂ = (2usin θ)/(g)·(2ucos θ)/(g) = 2/g·(u²sin 2θ)/(g) = 2R/g So t₁ t₂ ∝ R, answer (d).

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Comparing Two Projectiles

A projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ be the times of flights in the two cases, then the product of the two time of flights is proportional to

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Answer: (d)

For angles $\theta$ and $90^\circ - \theta$ : $t_1 t_2 = \frac{2u\sin\theta}{g}\cdot\frac{2u\cos\theta}{g}$
$= \frac{2}{g}\cdot\frac{u^2\sin 2\theta}{g} = \frac{2R}{g}$
So $t_1 t_2 \propto R$ , answer (d).

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