A rock is kicked horizontally off the edge of the cliff at a velocity of +15.0 m/s. If the canyon below is 80.0 m deep, how far from the edge of the cliff does the model rocket land?

60.6 m

Explanation:

For vertical displacement of 80 m , initial velocity downwards u = 0

h = ut + 1/2 g t²

80 = 0 + 1/2 x 9.8 x t²

t = 4.04 s

During this period , rock will be displaced horizontally with uniform velocity .

horizontal displacement = 15 x 4.04 = 60.6 m .

15.3 cos 65 degrees trig

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft