First slide

Projection Under uniform Acceleration

Question

The maximum range of a projectile is 500 m. If the particle is thrown up a plane is inclined at an angle of $30^{0}$ with the same speed, the distance covered by it along the inclined plane will be:

Moderate

A

250 m
B

500 m
C

750 m
D

100 m

Solution

For the maximum range, $θ = 45^{0}$
R = $\frac{u^{2} \sin 2 θ}{g} = \frac{u^{2}}{g} \sin 90^{0} = \frac{u^{2}}{g}$
or $500 = \frac{u^{2}}{g}$
The distance covered along the inclined plane can be obtained using the equation
$v^{2} - u^{2} = 2 as$
or 0- $u^{2} = 2 (- gsin 30^{0}) s$
or s = $\frac{u^{2}}{g} = 500 m$

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