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