Question

If a ball thrown vertically upward with some initial speed, it falls freely and reaches a
maximum height ‘h?. If its initial speed becomes double, what is its new maximum height?

Answer 1

Answer:

Explanation:

S = ut + ½at²

a = -g (since body is going up and acceleration is down so negative)

For maximum height

case 1

S = h1, u = u , t = u/g

h1 = u×u/g -½g×u²/g²

= u²/g-½u²/g = ½u²/g

Case 2

S = h2 u = 2u , t = 2u/g

h2 = 2u×2u/g -½g× 4u²/g²

= 4u²/g- 2u²/g

= 2u²/g

So,

h2/h1 = (2u²/g)/(½u²/g)

h2 = 4h1

So, new height is 4times the height of h