How long after being released does the stone strike the beach below the cliff? Answer in units of s.

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 12.6 m/s. The cliff is 24.4 m above a flat, horizontal beach. The acceleration of gravity is 9.8 m/s^2.

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What we must first understand is that the stone, if thrown horiztonally, has an intial velocity in the vertical direction of zero. Since you know the acceleration from gravity and the hieght that the stone is above the sand, you can use d = vit + 1/2at^2 solved algebraically for t.

Since Vi is zero, that portion of the equation drops out, you are then left with d = 1/2at^2 which you should then solve for t...

t = (2d/a)^.5 this will give you the appropriate flight time of the stone at 2.23 seconds.

The horizontal velocity imparted is not affected by gravity or it does not play a role in the time taken for the stone to land . It is only the gravitation that has a role in the time for the stone to reach the ground. The initial vertical velocity of the stone is 0 m/s and its horizontal velocity .

The vertical displacement , s = 24.4m. the depth of the cliff.

The law of motion of the stone relating to vertical displacement and time is: (1/2) gt^2 = s............(1)

g=9.8m/s^2 and s=24.4. Substituting the values g=9.8m/s^2 and s=24.4 in eq (1) we get :

(1/2)9.8t^2= 24.4 or

t= sqrt(2*24.4/9.8) = 2.2315 s is the time the stone takes to land.

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