A diver running 1.2 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.1 s later. How high was the cliff and how far from its base did the diver hit the water?
vertical drop
S = ut+(1/2)gt²
= 0 + (1/2)(9.8)*(3.1)^2
=47.089s
How far from base
1.2m/s * 3.1s = 3.72m
To find the height of the cliff, we can use the formula for vertical motion:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Given that the time taken by the diver to reach the water below is 3.1 seconds, we can substitute the values into the formula:
h = (1/2) * 9.8 * (3.1^2)
h = (1/2) * 9.8 * 9.61
h = 47.29 meters
Therefore, the height of the cliff is 47.29 meters.
To find the horizontal distance from the base of the cliff where the diver hits the water, we can use the formula for horizontal motion:
d = v * t
where d is the distance, v is the horizontal velocity, and t is the time.
Given that the diver is running horizontally with a velocity of 1.2 m/s and the time taken to reach the water is 3.1 seconds, we can substitute the values into the formula:
d = 1.2 * 3.1
d = 3.72 meters
Therefore, the diver hits the water approximately 3.72 meters from the base of the cliff.