Posted by **Cathy** on Sunday, February 1, 2009 at 11:07pm.

Problem: A Super Ball is thrown to the ground with an initial downward velocity of 0.70 m/s and pops back up to the hand of the thrower; the total time elapsed is 0.71 s. What height was the Super Ball thrown from?

Do I use the formula x = x0 + v0*t-0.5*g*t^2?

x0=0

v0=-0.70

t=0.71

g=9.81 m/s^2

- Physics -
**drwls**, Monday, February 2, 2009 at 5:06am
Yes, that is the correct equation to use.

1/2 of the 0.71s (0.355s) was spent going down and the other half going back up to the hand.

Let H be the initial height from which it was thrown. It starts at height H at t=0 and reaches height 0 at t - 0.355s

H - 0.70 * 0.355 - 9.81*(0.355)^2 = 0

- Physics (correction) -
**drwls**, Monday, February 2, 2009 at 5:18am
It reaches height 0 at t = 0.355s.

Note that v0 is negative (downward)

H - 0.70 * 0.355 - (9.81/2)*(0.355)^2

= 0

H = 0.249 + 0.618 = 0.867 m

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