A projectile can have the same range (R) for two angles of projection. Their initial velocities are same. If…

A projectile can have the same range (R) for two angles of projection. Their initial velocities are same. If $T_1$ and $\mathrm{T}_2$ are times of flight in two cases, then the product of two times of flight is directly proportional to
  1. $\frac{1}{R}$
  2. $\mathrm{R}^3$
  3. $\mathrm{R}^2$
  4. $R$

Solution

For same range, $R_1=R_2=R$ $\theta_1=\theta, \theta_2=\left(90^{\circ}-\theta\right)$ $\therefore \mathrm{T}_1=\frac{2 \mathrm{u} \sin \theta_1}{\mathrm{~g}}=\frac{2 \mathrm{usin} \theta}{\mathrm{g}}$ $\therefore \mathrm{T}_2=\frac{2 \mathrm{u} \sin \theta_2}{\mathrm{~g}}=\frac{2 \mathrm{u} \sin \left(90^{\circ}-\theta\right)}{\mathrm{g}}=\frac{2 \mathrm{u} \cos \theta}{\mathrm{g}}$ $\therefore \mathrm{T}_1 \cdot \mathrm{~T}_2=\left(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\right)\left(\frac{2 \mathrm{u} \cos \theta}{\mathrm{g}}\right)=\frac{2}{\mathrm{~g}}\left(\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{~g}}\right)=\frac{2 \mathrm{R}}{\mathrm{g}}$ $\therefore \mathrm{T}_1 \mathrm{~T}_2 \propto \mathrm{R}$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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