First slide

Projection Under uniform Acceleration

Question

A particle is projected under gravity with velocity $\sqrt{2 ag}$ 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:

Moderate

A

$\sqrt{(a^{2} + 1) h}$
B

$\sqrt{a^{2} h}$
C

$\sqrt{ah}$
D

$2 \sqrt{a (a + h)}$

Solution

Coordinate of point P are (R, - h)
Hence $R \leq 2 \sqrt{a (a + h)}$ $- h = R \tan θ - \frac{{gR}^{2}}{2 (2 ga)} (1 + \tan^{2} θ)$
or $R^{2} \tan^{2} θ - 4 aR \tan θ + (R^{2} - 4 ah) = 0$
For $θ$ to be real.
$(4 {aR}^{2}) \geq 4 R^{2} (R^{2} - 4 ah)$
or $4 a^{2} \geq (R^{2} - 4 ah)$
or $R^{2} \leq 4 a (a + h)$
or $R^{2} \leq 2 \sqrt{a (a + h)}$
$R_{\max} = 2 \sqrt{a (a + h)}$

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