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Battles

As shown in t...

Question

As shown in the figure, two projectiles are thrown from the top of a cliff of unknown height with equal initial speed $V_{0} = 10 m / s$ . One is projected at an angle $θ = 45^{\circ}$ above horizontal and second is projected at the same angle below horizontal. What is difference between range (in $m$ ) of the projectiles?

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Solution

Step 1: Find height for first particle.

Given,

Initial velocity $v_{0} = 10 m / s$
Projected at $θ = 45^{\circ}$
Suppose the height of cliff = $h$

For first particle, in vertical direction

$s = u t + \frac{1}{2} a t^{2}$

Taking the upward direction as positive and downward as negative.

$- h = V_{0}_{y} t_{1} + \frac{1}{2} a t_{1}^{2}$
$- h = V_{0} sin 45^{\circ} t_{1} - \frac{1}{2} g t_{1}^{2}$
$- h = \frac{V_{0}}{\sqrt{2}} t_{1} - \frac{1}{2} g t_{1}^{2} \dots (i)$

Step 2: Find height for second particle.

For second particle, in vertical direction

$s = u t + \frac{1}{2} a t^{2}$

Taking the upward direction as negative and downward as positive.

$+ h = \frac{V_{0}}{\sqrt{2}} t_{2} + \frac{1}{2} g t_{2}^{2} \dots (i i)$

Equation $(i)$ + equation $(i i)$ , we get –

$0 = \frac{V_{0}}{\sqrt{2}} (t_{1} + t_{2}) + \frac{1}{2} g (t_{2}^{2} - t_{1}^{2})$

$\frac{V_{0}}{\sqrt{2}} (t_{1} + t_{2}) + \frac{1}{2} g (t_{2} - t_{1}) (t_{1} + t_{2}) = 0$

$(t_{1} + t_{2}) (\frac{V_{0}}{\sqrt{2}} + \frac{g}{2} (t_{2} - t_{1})) = 0$

$t_{2} - t_{1} = - \frac{V_{0}}{\sqrt{2} \times \frac{g}{2}}$

$t_{2} - t_{1} = - \frac{10}{\sqrt{2} \times 5}$

$t_{1} - t_{2} = \sqrt{2} s e c \dots (i)$

Step 3: Find difference between range.

$R_{1} - R_{2} = V_{0} cos 45^{\circ} t_{1} - V_{0} cos 45^{\circ} t_{2}$

$= \frac{V_{0}}{\sqrt{2}} (t_{1} - t_{2})$ From $(i)$

$= \frac{V_{0}}{\sqrt{2}} \times \sqrt{2}$

$R_{1} - R_{2} = 10 m$
Final Answer: (10m)

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