question_answer

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.

A) \[2h\]                

B) \[3h\]

C) \[(5/3)h\]           

D) \[(4/3)h\]

Correct Answer: C

Solution :

[c] \[H=\frac{{{u}^{2}}}{2g}\]; given \[{{v}_{2}}=2{{v}_{1}}\]       (i) A to B :\[{{v}^{2}}_{1}={{u}^{2}}-2gh\]                   (ii) A to C : \[{{v}_{2}}^{2}={{u}^{2}}-2g(-h)\]   (iii) Solving (i),(ii),and (iii), we get the value of \[{{u}^{2}}\]as 10gh/3 and then we get the value of h by using \[H=\frac{{{u}^{2}}}{2g}\](see figure) or \[H=\frac{5h}{3}\]