If T is the total time of flight, h is the maximum height and R is the range for horizontal motion, the x and y coordinates of projectile motion and time t are related as

A

$y = 4 h (\frac{t}{T}) (1 - \frac{t}{T})$

B

$y = 4 h (\frac{x}{R}) (1 - \frac{x}{R})$

C

$y = 4 h (\frac{T}{t}) (1 - \frac{T}{t})$

D

$y = 4 h (\frac{R}{x}) (1 - \frac{R}{x})$

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Verified by Experts

The correct Answer is:A, B

a,b. $T = \frac{2 u sin θ}{g}, h = \frac{u^{2} {sin}^{2} θ}{2 g}, R = \frac{u^{2} sin 2 θ}{g}$
$y = u sin θ t - \frac{1}{2} {gt}^{2}$
$= 4 h t [\frac{u sin θ}{4 h} - \frac{gt}{8 h}]$
$= 4 h t [\frac{u sin θ}{4 (u^{2} {sin}^{2} θ / 2 g)} - \frac{gt}{8 (u^{2} {sin}^{2} θ / 2 g)}]$
$= 4 h t [\frac{1}{2 u sin θ / g} - \frac{t}{{(2 u sin θ / g)}^{2}}]$
$4 h t [\frac{1}{T} - \frac{t}{T^{2}}] = 4 h (\frac{1}{T}) (1 - \frac{t}{T})$
Now $x = u cos θ t, R = u cos θ T$
$\Rightarrow \frac{t}{T} = \frac{x}{R}, s o y = 4 h (\frac{x}{R}) (1 - \frac{x}{R})$ .

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Updated on:21/07/2023

Class 11 PHYSICS PROJECTILE MOTION

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