A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2 s later. How high
was the cliff and how far from its base did the diver hit the water?
work needed
work needed
1 answer:
Refer to the diagram shown below.
h = the height of the cliff.
d = the distance of the landing spot from the base of the cliff.
Assume that air resistance may be neglected, and g = 9.8 m/s².
The horizontal distance traveled in 2 seconds is
d = (1.8 m/s)*(2 s) = 3.6 m
The initial vertical velocity component is zero.
Therefore the vertical height traveled before landing on the water obeys the equation
h = (1/2)*(9.8 m/s²)*(2 s)² = 19.6 m
Answers:
The height of the cliff is 19.6 m
The distance of the landing spot from the base of the cliff is 3.6 m
h = the height of the cliff.
d = the distance of the landing spot from the base of the cliff.
Assume that air resistance may be neglected, and g = 9.8 m/s².
The horizontal distance traveled in 2 seconds is
d = (1.8 m/s)*(2 s) = 3.6 m
The initial vertical velocity component is zero.
Therefore the vertical height traveled before landing on the water obeys the equation
h = (1/2)*(9.8 m/s²)*(2 s)² = 19.6 m
Answers:
The height of the cliff is 19.6 m
The distance of the landing spot from the base of the cliff is 3.6 m
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