Asked by Sarah
A boy kicks a rock off a cliff with a speed of 15.5m/s at an angle of 53.5° above the horizontal. The rock hits the ground 5.37s after it was kicked. How high is the cliff?
I found the height of the cliff which is 7.45*10^1 m.
What is the speed of the rock right before it hits the ground?
I found the height of the cliff which is 7.45*10^1 m.
What is the speed of the rock right before it hits the ground?
Answers
Answered by
Damon
u = 15.5 cos 53.5= 9.22 m/s forever
Vi = 15.5 sin 53.5 = 12.5 initial
h = Hi + 12.5 t - 4.9 t^2
h = 0 at ground
Hi = 4.9 t^2 -12.5 t
Hi = 4.9(5.37)^2 - 12.5(5.37)
Hi = 82.1 meters
v = Vi -9.81 t
v = 12.5 - 9.81(5.37)
= - 40.2 m/s
Vi = 15.5 sin 53.5 = 12.5 initial
h = Hi + 12.5 t - 4.9 t^2
h = 0 at ground
Hi = 4.9 t^2 -12.5 t
Hi = 4.9(5.37)^2 - 12.5(5.37)
Hi = 82.1 meters
v = Vi -9.81 t
v = 12.5 - 9.81(5.37)
= - 40.2 m/s
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