A soccer ball is kicked from the edge of a cliff with a velocity of 12.0m/s horizontally. If the ball landed 47.9m from the base of the cliff, how tall is the cliff?
Use the formula to get the time:
x = vo,x * t
where
x = horizontal distance
vo,x = initial horizontal velocity
t = time
47.9 = 12 * t
t = 47.9 / 12
t = 3.99 s
Then use this formula to get the height of cliff:
h = vo,y * t - (1/2)gt^2
where
h = height
vo,y = initial vertical velocity
g = acceleration due to gravity = 9.8 m/s^2
Note that vo,y is zero.
h = 0 - (1/2)(-9.8)(3.99^2)
h = 78.0 m
hope this helps~ `u`
To find the height of the cliff, we can use the kinematic equations for motion. In this case, we only need to focus on the horizontal motion of the soccer ball since the vertical motion is not affected by the initial horizontal velocity.
The horizontal distance traveled by the ball is 47.9m, and the initial horizontal velocity is 12.0m/s. We can use the equation:
distance = velocity × time
Since the horizontal velocity remains constant, we know that the time the ball is in the air is the same as the time taken to cover the horizontal distance. Therefore, we can rearrange the equation to solve for time:
time = distance / velocity
Substituting the values, we get:
time = 47.9m / 12.0m/s
time = 3.99s (approximately)
Now, using the vertical motion, we can determine the height of the cliff. The vertical motion of the ball can be described using the equation:
height = initial vertical velocity × time + (1/2) × acceleration × time²
The initial vertical velocity is zero since the ball is not initially moving vertically, and the acceleration due to gravity is approximately -9.8m/s² (taking downward as negative). Rearranging the equation, we have:
height = (1/2) × acceleration × time²
Substituting the values, we get:
height = (1/2) × -9.8m/s² × (3.99s)²
height ≈ -76.94m
Since height cannot be negative, the magnitude of the cliff's height is approximately 76.94 meters.