Question

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s. How long does it take to reach the ground? How far does the rock land from the base of the cliff?

Answer 1

D 20202 hhsbdhhdhdhdhdhhs

Answer 2

d = 20.2 m

t = 2.02 s

Step-by-step explanation:

Horizontal Motion

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it hits the ground.

To calculate the time the object takes to hit the ground, we use the equation:

`\displaystyle t=\sqrt{(2h)/(g)}`

Note the time does not depend on the initial speed.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

`\displaystyle d=v.t`

The man standing on the edge of the h=20 m cliff throws a rock with a horizontal speed of v=10 m/s.

The time taken by the rock to reach the ground is:

`\displaystyle t=\sqrt{(2*20)/(9.8)}`

`\displaystyle t=√(4.0816)`

t = 2.02 s

Now for the horizontal range:

`\displaystyle d=10\cdot 2.02`

d = 20.2 m