First slide
Projection Under uniform Acceleration
Question

A smooth inclined plane is inclined at an angle q with the horizontal. A body starts from rest and slides down the inclined surface. Then the time taken by it to reach the bottom is

Easy
Solution

See fig.

Here F=mgsinθ or ma=mgsinθ

a=gsinθ   …(1)

Now, v = u + at = 0 + at 

or t = v/a   … (2)

But v=(2gh)   …(3)

t=(2gh)gsinθ=1sinθ2hg

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