Question

An old man standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 14 m/s. The cliff is 63.5 m above the water's surface.

a. How long does it take the stone to fall to the water?
b. With what speed does the stone strike the water?

Answer 1

a) t=3.6 sec

b)
`\mid v \mid =38\ m/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 trajectory ruled by a constant speed in the horizontal direction and a variable speed in the vertical direction, where the acceleration of gravity makes the object fall to the ground.

If we know the height h from which the object was launched, the time it takes to hit the ground is:

`\displaystyle t=\sqrt{\frac {2h}{g}}`

The horizontal speed is always constant:

vx=vo

But the vertical speed depends on the time and acceleration of gravity:

`vy=g\cdot t`

The magnitude of the velocity or final speed of the object is given by:

`\mid v \mid =√(vx^2+vy^2)`

a.

The stone was kicked over the cliff with a speed of v0=14 m/s, and the height it was thrown from is h=63.5 m, thus:

`\displaystyle t=\sqrt{\frac {2\cdot 63.5}{9.8}}`

t=3.6 sec

b.

The vertical speed is:

`vy=9.8\cdot 3.6`

vy=35.3 m/s

The final speed is calculated below:

`\mid v \mid =√(14^2+35.3^2)`

`\mid v \mid =√(1,442.09)`

`\boxed{\mathbf{\mid v \mid =38\ m/s}}`