Question

If the angle of projection of a projector with same initial velocity exceed or fall short of 45° by equal amount $α$ , then the ratio of horizontal ranges is [Kerala CEE 2014]

Moderate

Solution

Given, $θ_{1} = θ_{2} = 45^{0}$
$θ_{1} + θ_{2} = (45^{0} + α) + (45^{0} - α) = 90^{0}$
$u_{1} = u_{2} = u \Rightarrow R_{1} = R_{2} = ?$
$R_{1} = \frac{u_{1}^{2} \sin (θ_{1} + θ_{2})}{g}$
$\Rightarrow R_{1} = \frac{u^{2} \sin 90 °}{g} = \frac{u^{2}}{g}$ ……..(i)
Similarly, $R_{2} = \frac{u_{2}^{2} \sin (θ_{1} + θ_{2})}{g}$
$\Rightarrow R_{2} = \frac{u^{2} \sin 90 °}{g} = \frac{u^{2}}{g}$ ……..(ii)
The ratio of horizontal ranges
$\frac{R_{1}}{R_{2}} = \frac{u^{2} / g}{u^{2} / g} = 1 : 1 \Rightarrow R_{1} : R_{2} = 1 : 1$

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