Asked by Anonymous
A students on a balcony uses a slingshot to shoots rock straight up
at 29.4 m/s. If the rock takes 8.0 s seconds to hit the ground below the balcony, how high is the balcony above the ground?
at 29.4 m/s. If the rock takes 8.0 s seconds to hit the ground below the balcony, how high is the balcony above the ground?
Answers
Answered by
Henry
The velocity 0f the rock is zero at its' max. height:
Vf = Vo + gt = 0,
29.4 - 9.8t = 0,
-9.8t = -29.4,
t = -29.4 / -9.8 = 3s to reach max height
Time in flight = 8s.
Fall time = 8 - 3 = 5s,
d = 0*5 + 0.5*9.8*5^2,
d = 0 + 122.5 = 122.5m to gnd.
h = 122.5 - 44.1 = 78.4m.
= 4.9*3^2,
d = 88.2 - 44.1 = 44.1m up.
Vf = Vo + gt = 0,
29.4 - 9.8t = 0,
-9.8t = -29.4,
t = -29.4 / -9.8 = 3s to reach max height
Time in flight = 8s.
Fall time = 8 - 3 = 5s,
d = 0*5 + 0.5*9.8*5^2,
d = 0 + 122.5 = 122.5m to gnd.
h = 122.5 - 44.1 = 78.4m.
= 4.9*3^2,
d = 88.2 - 44.1 = 44.1m up.
Answered by
Henry
Oops!!
d (up) = Vo*t + 0.5*gt^2,
d (up) = 29.4*3 + 0.5*(-9.8)3^2,
d (up) = 88.2 - 44.1 = 44.1m.
d (up) = Vo*t + 0.5*gt^2,
d (up) = 29.4*3 + 0.5*(-9.8)3^2,
d (up) = 88.2 - 44.1 = 44.1m.
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