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Q.

A ball is thrown from the top of a tower in vertically upward direction. The velocity at a point h meter below the point of projection is twice of the velocity at a point h meter above the point of projection. Find the maximum height reached by the ball above the top of tower.

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a

2h

b

3h

c

(5/3)h

d

(4/3)h

answer is C.

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Detailed Solution

H=u22g; given v2=2v1    ......(i)A to B:v12=u2−2gh      .......(ii)A to C:v22=u2−2g(−h)    .......(iii)Solving (i), (ii) and (iii), we get the value of u2 as 10gh/3 and then we get the value of H by usingH=u22g (see figure) or H=5h3
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