A stone is thrown horizontaly from the top of a 25.00-m cliff. The stone lands at a distance of 40.00 m from the edge of the cliff. What is the horizontal velocity of the stone?
To find the horizontal velocity of the stone, we first need to determine the time it took for the stone to reach the ground.
We know the height of the cliff is 25.00 m. We can use the equation for free fall:
h = (1/2) * g * t^2
where:
h = height (25.00 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation, we get:
t^2 = (2 * h) / g
t^2 = (2 * 25.00) / 9.8
t^2 ≈ 5.102
Taking the square root of both sides:
t ≈ √5.102
t ≈ 2.259 s
Now that we know the time it took for the stone to fall, we can calculate the horizontal velocity.
Since the stone was thrown horizontally, the only force acting on it is due to gravity in the vertical direction. This force does not affect the horizontal motion of the stone. Therefore, the horizontal velocity remains constant throughout the motion.
The horizontal distance traveled, d, can be calculated using the equation:
d = v * t
where:
d = distance (40.00 m)
v = horizontal velocity
t = time (2.259 s)
Rearranging the equation, we get:
v = d / t
v = 40.00 / 2.259
v ≈ 17.69 m/s
Therefore, the horizontal velocity of the stone is approximately 17.69 m/s.