An object is dropped from a cliff and reaches a velocity of 47.5 m/s at the time of impact. a) How high is the cliff? b)How long did it take to hit the river below?
a) V = sqrt(2 g X)
Solve for X (the cliff height)
b) V = g*t
Solve for t
4.84 s
To answer these questions, we can use the equations of motion and the principles of free fall.
a) To find the height of the cliff, we need to use the equation that relates the final velocity, initial velocity, acceleration, and displacement:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (47.5 m/s)
vi = initial velocity (0 m/s, since the object is dropped)
d = displacement (height of the cliff, which is what we need to find)
a = acceleration (which is the acceleration due to gravity, approximately 9.8 m/s^2)
Rearranging the equation, we have:
d = (vf^2 - vi^2) / (2a)
Substituting the values:
d = (47.5^2 - 0) / (2 * 9.8)
d = (2256.25) / 19.6
d ≈ 115.10 meters
Therefore, the height of the cliff is approximately 115.10 meters.
b) To find the time it takes to hit the river, we can use the equation that relates initial velocity, final velocity, acceleration, and time:
vf = vi + at
Where:
vf = final velocity (47.5 m/s)
vi = initial velocity (0 m/s, since the object is dropped)
a = acceleration (which is the acceleration due to gravity, approximately 9.8 m/s^2)
t = time (which is what we need to find)
Rearranging the equation, we have:
t = (vf - vi) / a
Substituting the values:
t = (47.5 - 0) / 9.8
t ≈ 4.85 seconds
Therefore, it took approximately 4.85 seconds for the object to hit the river below.