Question

A ball is shot vertically upward with a given initial velocity. It reaches a maximum height of 100 m.
If on a second shot, the initial velocity is doubled then the ball will reach a maximum height of​

Answer 1

Dear Friend,

◆ Answer -

s = 400 m

● Explanation -

Let u be initial velocity of the ball after hitting & v be final velocity of the ball at the top.

For first attempt,

v^2 = u^2 + 2as

0^2 = u^2 + 2(-10)(100)

u^2 = 2000

u = √2000 m/s

For second attempt,

v'^2 = u'^2 + 2as'

0^2 = (2×√2000)^2 + 2×(-10)×s

s = 8000 / (20)

s = 400 m

Therefore, ball will reach a maximum height of 400 m in second shot.

Thanks dear.

Answer 2

Answer:◆ Answer -

s = 400 m

● Explanation -

Let u be initial velocity of the ball after hitting & v be final velocity of the ball at the top.

For first attempt,

v^2 = u^2 + 2as

0^2 = u^2 + 2(-10)(100)

u^2 = 2000

u = √2000 m/s

For second attempt,

v'^2 = u'^2 + 2as'

0^2 = (2×√2000)^2 + 2×(-10)×s

s = 8000 / (20)

s = 400

Explanation: