Q.

A particle is projected with a speed v0 at an angle θ above the horizontal surface such that the ratio of the kinetic energy at the highest point and the point of projection is 3:4. The change in velocity of the particle between these two points is

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a

(v0/2) vertically downward.

b

(v0/2) vertically upward.

c

3v0 due east.

d

3v0/2 vertically downward.

answer is A.

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Detailed Solution

The ratio of the kinetic energy at the highest point and point of projection isKHK0=34⇒(1/2)mv0cos⁡θ2(1/2)mv02=34⇒cos2⁡θ=34⇒θ=π6Change in velocity =v0sinθ=v0/2.
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