Question: An object is thrown vertically upward and returns after 10 seconds. Find the initial velocity and its maximum height.

Solution:

Total time of upward and downward motion = T_{tot} = 10 s.

So, time for upward motion = T

_{tot}/ 2 = 5 s.So it takes 5 s to reach the maximum height.

Say, its initial velocity is u. At the highest point v = 0

Using the motion equation for the vertical upward motion: v=u-gt

In this case, at maximum height, this equation becomes

0 = u – 9.8 x 5

=> u = 49 m/s

Initial Velocity = 49 m/s [ Answer]

Maximum height = H_{max}= u^{2}/(2g) = 49^{2}/(2×9.8) m = 122.5 m[Answer]