First slide

Projection Under uniform Acceleration

Question

Equation of trajectory of a particle projected from a point on ground is $y = (x - \frac{x^{2}}{240})$ where x and y are in metre. If $g = 10 m / s^{2}$ , velocity of projection is

Moderate

A

$20 \sqrt{6} m / s$
B

$10 \sqrt{2} m / s$
C

30 m/s
D

$20 \sqrt{3} m / s$

Solution

$y = x (1 - \frac{x}{240})$
Comparing this equation with the standard equation
$y = xtanα (1 - \frac{x}{R}), We get$
$tanα = 1 \Rightarrow α = 45^{0}$ and Range = 240 m
$Now \frac{u^{2}}{g} \sin 2 α = 240 \Rightarrow \frac{u^{2}}{10} \sin (2 \times 45^{0}) = 240 \Rightarrow u = 20 \sqrt{6} m / s$

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