A ball is thrown vertically upward from the edge of a cliff of height 131 m. The initial velocity is 77.7 m/s. how high is the cliff?
131 m. :)
To determine the height of the cliff, we need to analyze the motion of the ball.
The given information includes:
- Initial velocity (u) = 77.7 m/s (upward)
- Final velocity (v) = 0 m/s (at maximum height)
- Distance (s) = ? (height of the cliff) = 131 m (upward)
Understanding the motion, we can use the equation for displacement (s):
s = ut + (1/2)at^2
Where:
- s is the displacement or height of the cliff
- u is the initial velocity
- t is the time taken
- a is the acceleration due to gravity (which is -9.8 m/s^2 since the ball is moving upward)
First, we need to find the time taken to reach the maximum height. In this case, the final velocity is 0 m/s, so we can use the equation:
v = u + at
0 = 77.7 - 9.8t
Simplifying the equation:
9.8t = 77.7
t = 77.7 / 9.8
t ≈ 7.93 seconds
Now that we have the time, we can substitute the values into the displacement equation:
s = ut + (1/2)at^2
s = 77.7 * 7.93 + (1/2) * (-9.8) * (7.93^2)
s ≈ 131 meters
Therefore, the height of the cliff is approximately 131 meters.