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leke10 · Oct 6, 2010, 11:03 PM

a ball is thrown straight up and rises to a maximum height of 22m. At what height is the speed of the ball equal to half of its initial value?assume the ball starts at a height 2.2m above the ground

Unknown008 · Oct 7, 2010, 02:51 AM

Find the initial speed first.

Use:

$v^2 = u^2 + 2as$

Where v is the final velocity
u the initial velocity
a the acceleration
s the displacement.

We'll take up as positive.

$0^2 = u^2 + 2(-9.8)(22-2.2)$

$u = 19.7\ m/s$

So, now, you need to use the same formula, but with the final speed v as being half that speed you just found, that is 19.7/2 = 9.85 m/s.

$(9.85)^2 = (19.7)^2 + 2(-9.8)s$

Find s.

This s is the displacement above the starting height. So, add 2.2 m to this.

Post what you get! :)

ANS: 14.85 m + 2.2 m = 17.05 m