# Physics

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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 -

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) -

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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