vectors and projectile motion

Question

A rock is thrown at 20m/s horizontally off a cliff 100m high.How far away from the cliff does the rock land?

Answer 1

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

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

follow in the absence of gravity.

- A vertical fall due to gravity, which is what the rock 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 rock after time t.

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

x(t) = vt

y(t) = h - gt²/2,

where

v is the initial horizontal velocity

h is the height of the cliff

g is the gravitational acceleration

The fall continues until the rock 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 rock got horizontally by:

x(t₁) = vt₁ = v√(2h/g).

Substituting actual values

v = 20m/s

h = 100m

g = 9.8m/s²

x(t₁) = 20 m/s √(2 x 100 m / 9.8 m/s²) = 90.35 m