Question

A ball is thrown vertically upward. The velocity at one-third of maximum height from ground is 8 m/s. The velocity with which the ball was thrown is
[Take g = 10 m/s?]
(1) 4√2 m/s (2) 4√3 m/s
(3) 4√6 m/s (4) 2√6 m/s​

Answer 1

Answer:

13.85 m/s

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.

Answer 2

Answer:

$4\sqrt{6 }$

Explanation:

At total Height H

v=0

therefore

${u}^{2} = 2gh \: - (1)$

Now At height H/3

v=8

${v}^{2} = {u}^{2} - \frac{2gh}{3} \\ 64 = {u}^{2} - \frac{ {u}^{2} }{3} \\ u = \sqrt{3 \times 8 \times 4} \\ u = 4 \sqrt{6}$

here using equation (1) we can replace 2gh with u^2