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Question

A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground?

Answer 1

Do you refer to its speed or velocity? Its velocity has a horizontal and a vertical component. When he throws it horizontally, its vertical component will be exactly the same as if he just dropped it. Its horizontal component will be 10.0 m/s.
To calculate the vertical velocity component, use these equations:
For s = distance covered (displacement)
u = intitial velocity
v = final velocity
a = acceleration (due to gravity - 9.8 ms^-2)

Use v^2 = u^2 + 2as
In the first case, v = 30, u = 10, a = 9.8
Substitute in these numbers, and you get:
900 = 100 + 19.6s
Reaarange this, gives: (900-100)/19.6 = s = 40.8 m.

In the second case then you have ascertained that s = 40.8,
u = 0, a = 9.8
v^2 = u^2 + 2as
Substitute in the numbers:
v^2 = 0 + 800
v = square root 800 = 28.2 m/s

Answer 2

It should still be 30 metres per second - as it is the vertical component of motion that is providing the acceleration, it doesn't matter that he is throwing it sideways.