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=u22g ; given v2=2v1

(i) A to B : v12=u22gh

(ii) A to C : v22=u22g(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 using

H=u22g or H=5h3



NCERT Chapter Video Solution

Dual Pane