A stone is thrown horizontally at a speed of 4.8 m/s from the top of a cliff 88.5 m high.
How long does it take the stone to reach the bottom of the cliff?
u = 0
s = 88.5 m
a = 9.8 m/s^2
s = ut + 1/2 at^2
88.5 = o + 1/2 9.8 * t^2
= 4.9 * t^2
t^2 = 88.5 / 4.9
= 18.06
t = 4.25
Horizonal displacement = vt
= 4.8 * 4.5
=21.6 m
To find the time it takes for the stone to reach the bottom of the cliff, we can use the kinematic equation:
h = v*t + (1/2)*a*t^2
Where:
h = height of the cliff (88.5 m)
v = initial vertical velocity (0 m/s, since the stone is thrown horizontally)
t = time taken
a = acceleration due to gravity (-9.8 m/s^2, since gravity pulls the stone downward)
Since the stone is thrown horizontally, there is no acceleration in the horizontal direction.
First, we need to find the time it takes for the stone to fall from the top of the cliff to the ground. We can ignore the horizontal speed of the stone for this calculation.
Using the equation:
h = (1/2)*a*t^2
Substituting the values:
88.5 = (1/2)*(-9.8)*t^2
To solve for t^2, we can rearrange the equation:
t^2 = (2*88.5) / 9.8
t^2 = 18
Taking the square root of both sides, we get:
t = √18
t ≈ 4.24 s
Therefore, it takes approximately 4.24 seconds for the stone to reach the bottom of the cliff.