Asked by JJ
You lean over the edge of a building that is 106 meters above the ground and throw an object straight upwards. This object takes 3 seconds to reach its maximum vertical displacement above your hand and then falls to the ground. From the moment it leaves your hand, how long does the object take to travel to the ground in seconds?
Answers
Answered by
Henry
Vf = Vo + gt,
Vo = Vf - gt,
Vo = 0 - (-9.8)3 = 29.4m/s.
h = 106 + (Vf^2 - Vo^2) / 2g,
h=106 + (0 - (29.4)^2) / -19.6=150.1m
above ground.
d = Vo*t + 4.9t^2 = 150.1m,
0 + 4.9t^2 = 150.1,
t^2 = 30.63,
t(dn) = 5.53s,
T = t(up) + t(dn) = 3 + 5.53 = 8.53s in flight.
Vo = Vf - gt,
Vo = 0 - (-9.8)3 = 29.4m/s.
h = 106 + (Vf^2 - Vo^2) / 2g,
h=106 + (0 - (29.4)^2) / -19.6=150.1m
above ground.
d = Vo*t + 4.9t^2 = 150.1m,
0 + 4.9t^2 = 150.1,
t^2 = 30.63,
t(dn) = 5.53s,
T = t(up) + t(dn) = 3 + 5.53 = 8.53s in flight.
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