A stone is thrown horizontally 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 initial horizontal velocity of the stone?
15.60m/s
Engr joe
how to do this question???
To find the initial horizontal velocity of the stone, we can make use of the fact that the vertical motion of the stone does not affect its horizontal motion.
Given:
- The vertical displacement (height) of the stone, h = 25.00 m
- The horizontal displacement (distance) of the stone, d = 40.00 m
We can use the formula for horizontal motion to find the initial horizontal velocity (Vx).
The formula for horizontal motion is:
d = Vx * t
Where:
- d is the horizontal displacement of the stone
- Vx is the initial horizontal velocity of the stone
- t is the time it takes for the stone to travel the horizontal distance
Since the stone is thrown horizontally, the time it takes to reach the horizontal distance, t, is the same as the time it takes for the stone to fall vertically. We can find this time using the formula for vertical motion.
The formula for vertical motion is:
h = (1/2) * g * t^2
Where:
- h is the vertical displacement (height) of the stone
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time it takes for the stone to fall vertically
Rearranging the formula for vertical motion to solve for t:
t^2 = (2 * h) / g
t = sqrt((2 * h) / g)
Now, plugging in the given values:
h = 25.00 m
g = 9.8 m/s^2
t = sqrt((2 * 25.00 m) / 9.8 m/s^2)
t = sqrt(5.1 s^2)
Now that we have the value of t, we can find the initial horizontal velocity, Vx.
d = Vx * t
Rearranging the formula for horizontal motion to solve for Vx:
Vx = d / t
Plugging in the values:
d = 40.00 m (the horizontal displacement of the stone)
t = sqrt(5.1 s^2) (the time it takes for the stone to fall vertically)
Vx = 40.00 m / sqrt(5.1 s^2)
Calculating this expression will give us the initial horizontal velocity of the stone.