Question

You toss a ball straight up, and it reaches a maximum height h above the launch position. If you want to double the length of the time interval during which the ball stays in the air above the launch position, to what maximum height do you have to throw it? Express your answer in terms of h.

Answer 1

The maximum height you have to throw it will be 4h

What is a Projectile ?

A projectile is any object thrown into the space or atmosphere which take a trajectory path.

If you toss a ball straight up, and it reaches a maximum height h above the launch position, Then the maximum height reach will be

v² = u² - 2gh

But at maximum height, v = 0

u² = 2gh

h = u² / 2g ...... (1)

Also, the time taken will be;

v = u - gt

since v = 0,

u = gt

Substitute u into equation 1

h = (gt)² / 2g

If you want to double the length of the time interval during which the ball stays in the air above the launch position, then,

u = g2t and the maximum height will be

H = (g2t)² / 2g

H = 4g²t² / 2g

H = 4(gt)² / 2g

but h = (gt)² / 2g

H = 4h

Therefore, the maximum height you have to throw it will be 4h Expressed in terms of h.

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