Standing on the edge of the roof of a building, you throw a stone upwards with an initial speed of 6.47 m/s. Then the stone falls to the ground, 10.3m below the point where the stone leaves your hand. How fast does the stone hit the ground? How long does the stone stay in the air? Ignore air resistance and take g = -9.8 m/s2.
vy = v° – g * t
peak when vy = 0 —- gt = v° — gt = 6.47 — t = 6.47 / 9.8 = 0.66 s
average peak height vt = (6.47 + 0) /2 0.66 = 2.13576 m
free fall from height 10.03 + 2.13576 = 12.16576 m
e = 1/2 gt^2
t^2 = 2 e / g = 24.3315 / 9.8 = 2.4828 —- t = √ 2.4828 = 1.5757 s
speed at which the stone hits the ground v = gt = 9.8 1.5757 = 15.4418 m/s
stone in the air 0.66 s + 1.5757 = 2.2357 s
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.ht…
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