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.
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?
2 answers
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
2 answers