You are standing at the top of a cliff that has a stairstep configuration. There

Question

You are standing at the top of a cliff that has
a stairstep configuration. There is a vertical
drop of 5 m at your feet, then a horizontal
shelf of 8 m, then another drop of 5 m to the
bottom of the canyon, which has a horizontal
floor. You kick a 0.97 kg rock, giving it an
initial horizontal velocity that barely clears
the shelf below.
8 m
x
10 m
v
5 m
5 m
The acceleration of gravity is
9.8 m/s2 . Consider air friction to be negligi-
ble.How far from the bottom of the second cliff
will the projectile land?
Answer in units of m

Explanation / Answer

Assuming no air resistance, the only force acting on the rock is gravity.

In the x-direction, its position w.r.t. time is

x(t) = v * t

In the y-direction

y(t) = - 1/2 * a * t2

Re-arranging

y(x) = - 4.9 * x2 / v2

We know that, when y = -5m, x = 8m.

Therefore

-5 = -4.9 * 64 * v2

v = 0.13 m/s

Thus, when the rock is at the bottom of the canyon

-10 = -4.9 * 0.0169 * x2

x = 11m

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