**Problem 10:** Calculate the angle of projection for which
K.E at the highest point of its trajectory is equal to one fourth of its K.E at
the point of projection.

**Solution**

**Note:** At the angle of projection, the initial
velocity has two components, v_{x} = v_{i}cosθ and v_{y}
= v_{i}sinθ . At the highest point, the y-component of the velocity
becomes 0 and the x-component undergoes no change.

**Given:** K.E at the point of projection = 4× K.E at
the highest point

Now K.E at the point of projection

And K.E at the highest point

Now according to the condition of the question,

Therefore, angle of
projection is 60^{0}.

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