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
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
2 answers
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
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
1 answer