First slide

Projection Under uniform Acceleration

Question

A body is projected with a velocity $20 \sqrt{6} m / s$ from a point P on an inclined plane having inclination 30⁰ with horizontal making an angle of 45⁰with the plane. Then the time after which the body will strike the inclined plane is

Moderate

A

4 sec
B

6 sec
C

7 sec
D

8 sec

Solution

Component of initial velocity normal to the inclined plane u = $20 \sqrt{6} \sin 45^{0} m / s = 20 \sqrt{3} m / s$ .
Using the equation of time of flight we can write , $T = \frac{2 u}{g \cos 30^{0}} = \frac{2 \times 20 \sqrt{3}}{10 \times \frac{\sqrt{3}}{2}} s = 8 s$

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