Question

a ball is thrown vertically upward. the velocity at one third of maximum height from ground is 8m/sec. the velocity with which the ball thrown is (take g=10m/s')

Answer 1

using mechanical energy conservation.
1/2mv1^2 =mgh+ 1/2mv2^2
v1 = initial speed
v2= speed at h/3
h=H/3
where H is the maximum height

Answer 2

Answer:

Explanation:

Velocity of maximum height = 1/3 (Given)

Maximum height from ground = 8m/s (Given)

Let the velocity with the ball is thrown upwards be = v m/s.

Thus, then the maximum height achieved will be -  

= H = v²/2g

At one-third of max. height v = 8 m/s, therefore -  

= H/3 =64/2g

= H =32 ×3/g  

= 9.6 m

v =√(2gH)  

= √(2 × 10 ×9.6)

= √(192)  

= 13.85 m/s

Thus, The velocity with which the ball was thrown is 13.85 m/s.