Kinematics Question 174
Question: 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.
Options:
A) $ 2h $
B) $ 3h $
C) $ (5/3)h $
D) $ (4/3)h $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ H=\frac{u^{2}}{2g} $ ; given $ v _{2}=2v _{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} $ or $ H=\frac{5h}{3} $
BETA
