Posted by **mb** on Thursday, September 25, 2008 at 11:22pm.

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18 m/s. the cliff is 50 m above a flat, horizontally beach. how long after being released does the stone strike the beach below the cliff?

- physics -
**DrBob222**, Thursday, September 25, 2008 at 11:28pm
distance = 1/2*g*t^2

solve for t. Check m thinking.

- physics -
**Imalka**, Friday, August 10, 2012 at 6:28am
We can talk about this projection as two simultaneous and perpendicular motion.

Assuming that the height of the student can be omitted, the gravity is a constant with a value of 10m/s, no friction or disturbance in air and the Earth is an inertial frame:

The time of flight can be gained by using S = ut + (1/2)*at^2 downwards and as the initial value for the vertical velocity is 0 we can omit the 'ut' part.

50 = (1/2)*10t^2

t = √10 s

The horizontal distance can be gained through S = ut (for the rock flies with a uniform horizontal velocity and with no horizontal acceleration).

S = 18*√10

S = 18√10 m

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