Asked by anonymous
A rock is thrown off a vertical cliff at an angle of 58.6o above horizontal. The cliff is 84.6 m high and the ground extends horizontally from the base. The initial speed of the rock is 21.8 m/s. Neglect air resistance.
To what maximum height, in meters, does the rock rise above the edge of the cliff?
To what maximum height, in meters, does the rock rise above the edge of the cliff?
Answers
Answered by
bobpursley
vertical speed initial :(Vi): 21.8*sin58
initial KE in vertical: 1/2 m (21.8*sin58)^2
final change in potential energy:
mgh=1/2 m (21.8*sin58)^2
solve for height h above the upper edge of the cliff.
initial KE in vertical: 1/2 m (21.8*sin58)^2
final change in potential energy:
mgh=1/2 m (21.8*sin58)^2
solve for height h above the upper edge of the cliff.
Answered by
anonymous
ok i got! what about these questions:
How high above the ground, in meters, is the rock 2.21 s before it hits the ground?
How far horizontally from the cliff, in meters, is the rock 2.21 s before it hits the ground?
using the same numbers given above
How high above the ground, in meters, is the rock 2.21 s before it hits the ground?
How far horizontally from the cliff, in meters, is the rock 2.21 s before it hits the ground?
using the same numbers given above
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