Question
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?
Answers
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`
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`