Question

A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff. Determine the magnitude of the launch velocity

Answer 1

`v_o=40.14\ m/s`

Step-by-step explanation:

Horizontal Launch

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

• The horizontal speed is constant and equal to the initial speed
  `v_o`
• The vertical speed is zero at launch time, but increases as the object starts to fall
• The height of the object gradually decreases until it hits the ground
• The horizontal distance where the object lands is called the range

We have the following formulas

`\displaystyle v_x=v_o`

`\displaystyle x=v_o.t`

`\displaystyle v_y=g.t`

`\displaystyle y=(gt^2)/(2)`

Where
`v_o` is the initial horizontal speed,
`v_y` is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

`\displaystyle y=(gt^2)/(2)`

Rearranging and solving for t

`\displaystyle 2y=gt^2`

`\displaystyle t^2=(2\ y)/(g)`

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

We then replace this value in

`\displaystyle x=v_o.t`

To get

`\displaystyle v_o=(x)/(t)`

`\displaystyle v_o=\frac{x}{\sqrt{(2y)/(g)}}`

`\displaystyle v_o=\sqrt{(g)/(2y)}.x`

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing
`v_o`

`\displaystyle v_o=\sqrt{(9.8)/(2(35.2))}\ 107.6`

The launch velocity is

`\boxed{v_o=40.14\ m/s}`