Asked by Ashley
You are driving off of a horizontal cliff, and your car has a horizontal velocity of 29.33 m/s at the instant you leave the cliff. The cliff is an unknown height above a rocky ravine, but you land 146.8 m away from the sheer cliff face.
How tall is the cliff?
How tall is the cliff?
Answers
Answered by
MathMate
Horizontal velocity does not change in a free-fall, so
time = distance/velocity
= 146.8/29.33
= 5.0051 s. (keeping 5th significant fig.)
Initial vertical velocity = 0 for free fall, so
H=(1/2)at^2
=(1/2)*(-9.81)(5.0051^2)
=-122.9 m
Ans. the ground is 122.9m below the cliff.
time = distance/velocity
= 146.8/29.33
= 5.0051 s. (keeping 5th significant fig.)
Initial vertical velocity = 0 for free fall, so
H=(1/2)at^2
=(1/2)*(-9.81)(5.0051^2)
=-122.9 m
Ans. the ground is 122.9m below the cliff.
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