An object is projected horizontally at 8.0 m/s from the top of a 122.5 m cliff. how far from the base of the cliff will the obje
ct strike the ground?
1 answer:
Refer to the diagram shown below.
Assume g = 9.8 m/s² and neglect air resistance.
All quantities are measured as positive upward.
The vertical launch velocity is u = 0.
Let t = the time of flight.
Use the formula s = ut + (1/2)at²
Then
(-122.5 m) = 0 + (1/2)*(-9.8 m/s²)*(t s)²
t² = 122.5/4.9 = 25
t = 5 s
The horizontal distance traveled is
d = (8.0 m/s)*(5 s) = 40 m
Answer: 40 m
Assume g = 9.8 m/s² and neglect air resistance.
All quantities are measured as positive upward.
The vertical launch velocity is u = 0.
Let t = the time of flight.
Use the formula s = ut + (1/2)at²
Then
(-122.5 m) = 0 + (1/2)*(-9.8 m/s²)*(t s)²
t² = 122.5/4.9 = 25
t = 5 s
The horizontal distance traveled is
d = (8.0 m/s)*(5 s) = 40 m
Answer: 40 m
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