Asked by Zack
The road runner drops a boulder from a 53 meter high building. When the block is 14 meters above the ground, the 2 meter tall Wiley Coyote looks up and sees the boulder. How much time does he have to get out of the way?
Answers
Answered by
Henry
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*(53-14) = 764.4
V = 27.65 m/s at 14 m above gnd.
h = Vo*t + 0.5g*t^2 = 14-2 = 12
27.65t + 0.5g*t^2 = 12
4.9t^2 + 27.65t - 12 = 0
Use Quadratic Formula and get:
t = 0.405 s. To get out of the way.
V^2 = 0 + 19.6*(53-14) = 764.4
V = 27.65 m/s at 14 m above gnd.
h = Vo*t + 0.5g*t^2 = 14-2 = 12
27.65t + 0.5g*t^2 = 12
4.9t^2 + 27.65t - 12 = 0
Use Quadratic Formula and get:
t = 0.405 s. To get out of the way.
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