Question: An object is thrown vertically upward and returns after 10 seconds. Find the initial velocity and its maximum height.
Total time of upward and downward motion = Ttot = 10 s.
So, time for upward motion = Ttot / 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 = Hmax= u2/(2g) = 492/(2×9.8) m = 122.5 m [Answer]