A girl on the edge of a vertical cliff 7.6 m high throws a stone horizontally away from the cliff with a speed of 6 m/s. How far from the base of the cliff does it land?

Answer 1

The stone follows a parabola resulting from a combination of two movements:

- A horizontal trajectory due to inertia, which is what the stone would

follow in the absence of gravity.

- A vertical fall due to gravity, which is what the stone would follow

in the absence of any initial horizontal velocity.

You can view the trajectory in an X-Y coordinate space, where

the X coordinate is along the land, and

the Y coordinate is flush against the edge of the cliff.

Then [x(t), y(t)] is the position of the stone after time t.

(We start measuring time t from the instant the stone leaves the cliff.)

x(t) = vt

y(t) = h - gt²/2,

where

v = 6 m/s is the initial horizontal velocity

h = 7.6 m is the height of the cliff

g = 9.81 m/s² is the gravitational acceleration

The fall continues until the stone hits the land at some time t₁,

which we can calculate because at time t₁, y(t₁) = 0. That is,

h - gt₁²/2 = 0

t₁ = √(2h/g).

Having obtained t₁, we can calculate how far the stone got horizontally by:

x(t₁) = vt₁ = v √(2h/g)

Substituting actual numbers

x(t₁) = 6×√(2×7.6/9.81) = 7.47 m