A stone is thrown horizontally at a speed of 11.6 m/s from the top of a cliff 77.2 m high.?
How long does it take the stone to reach the bottom of the cliff?
How far from the base of the cliff does the stone hit the ground?
What is the vertical?

answers:
oubaas say: A stone is thrown horizontally at a speed of 11.6 m/s from the top of a cliff 77.2 m high.
How long (t) does it take the stone to reach the bottom of the cliff?
h = Voy*t+g/2*t^2
since the ball is thrown horizontally ,then Voy = 0 and the equation above becomes :
h = g/2*t^2
which leads to :
t = √2h/g = √77.2*2/9.806 = 3.97 sec
How far (d) from the base of the cliff does the stone hit the ground?
d = Vox*t = 11.6*3.97 = 46.0 m
What is the vertical speed Vy?
Vy = g*t = 9.806*3.97 = 38.9 m/sec
what is the speed V ?
V = √Vox^2+2gh = √11.6^2+154.4*9.806 = 40.6. m/sec

electron1 say: To determine the time for the stone to reach the bottom of the cliff, use the following equation.
d = vi * t + g * t^2
Since the stone is thrown horizontally, its initial vertical velocity is 0 m/s.
77.2 = ½ * 9.8 * t^2
t = √(77.2 ÷ 4.9)
The time is approximately 4 seconds.
How far from the base of the cliff does the stone hit the ground?
d = vh * t
d = 11.6 * √(77.2 ÷ 4.9)
This is approximately 46 meters.
What is the vertical?
To determine the stone’s vertical velocity just before it hits the ground, use the following equation.
vf = vi + g * t, vi = 0
vf = 9.8 * √(77.2 ÷ 4.9)
This is approximately 39 m/s.
I hope this is helpful for you.

Andrew Smith say: see https://answers.yahoo.com/question/index... for the method

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