Question

A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30 m/s. if the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10 m/s, how fast would the stone be traveling just before it hits the ground

Answer 1

The velocity just before hitting the ground is
`v_f = 30 m/s`

Step-by-step explanation:

From the question we are told that

The initial speed is
`u = 10 m/s`

The final speed is
`v = 30 \ m/s`

From the equations of motion we have that

`v^2 =u^2 + 2as`

Where s is the distance travelled which is the height of the cliff

So making it the subject of the the formula we have that

`s = (v^2 - u^2 )/(2a)`

Where a is the acceleration due to gravity with a value
`a = 9.8m/s^2`

So

`s = (30^2 - 10^2 )/(2 * 9.8 )`

`s = 40.8 \ m`

Now we are told that was through horizontally with a speed of

`v_x =10 m/s`

Which implies that this would be its velocity horizontally through out the motion

Now it final velocity vertically can be mathematically evaluated as

`v_y = √(2as)`

Substituting values

`v_y = √((2 * 9.8 * 40.8))`

`v_y = 28.3 \ m/s`

The resultant final velocity is mathematically evaluated as

`v_f = √(v_x^2 + v_y^2)`

Substituting values

`v_f = √(10^2 + 28.3^2)`

`v_f = 30 m/s`