A particle is projected under gravity with velocity $\sqrt{2 a g}$ from a point at a height $h$ above the level plane at an angle $θ$ to it. The maximum range $R$ on the ground is $\frac{x \sqrt{a (a + h)}}{4}$ . Find $x$

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Solution

Step 1: Draw rough diagram for the particle motion.
Step 2: Find maximum range R.
Given, particle is prjected with velocity,
$u = \sqrt{2 a g}$
When the body is projected at an angle $θ$ with horizontal in the upward direction from a height $h$ , then
$h = (- u sin θ) T + \frac{1}{2} g T^{2}$
And time of flight, $T = \frac{R}{u cos θ}$
$\Rightarrow h = (- u sin θ) (\frac{R}{u cos θ}) + \frac{1}{2} g {(\frac{R}{u cos θ})}^{2}$
$h = - R tan θ + \frac{g R^{2}}{4 a g} ({sec}^{2} θ)$
$h = - R tan θ + \frac{R^{2}}{4 a} (1 + {tan}^{2} θ)$

$\Rightarrow R^{2} {tan}^{2} θ - 4 a R tan θ + (R^{2} - 4 a h) = 0$

For $θ$ to be real

$(4 a R)^{2} \geq 4 R^{2} (R^{2} - 4 a h)$

$\Rightarrow 4 a^{2} (R^{2} - 4 a h) \Rightarrow R^{2} \leq 4 a (a + h)$

$\Rightarrow R \leq 2 \sqrt{a (a + h)}$

$R_{m a x} = 2 \sqrt{a (a + h)}$
Compare it with , $R = \frac{x \sqrt{a (a + h)}}{4}$
$\Rightarrow x = 8$
Final answer: 8

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A particle is projected under gravity with velocity $\sqrt{2 a g}$ from a point at a height $h$ above the level plane at an angle $θ$ to it. The maximum range $R$ on the ground is $\frac{x \sqrt{a (a + h)}}{4}$ . Find $x$