A diver running 2.0 m/s dives outhorizontally from an edge of a vertical cliff and 2.6 s later reaches the water below. How high was the cliff? How far from its base did the diver hit the water?
How high was the cliff?
How far will an object fall in 2.6 s ?
(1/2)g t^2 = ___
The horizontal distance travelled is the running (horizontal) velocity (2.0 m/s) mu;ltiplied by the 2.6 s fall time.
That would be ___ meters
43.776
To determine the height of the cliff and the horizontal distance the diver traveled, we can use the equations of motion.
First, let's find the height of the cliff. We can use the equation:
h = ut + (1/2)gt^2
Where:
h is the height of the cliff,
u is the initial vertical velocity of the diver (which is 0, since the diver is starting from rest at the edge of the cliff),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
and t is the time the diver takes to reach the water (2.6 s).
Plugging in the values:
h = (0)(2.6) + (1/2)(9.8)(2.6)^2
h = 0 + (1/2)(9.8)(6.76)
h = 0 + 49(6.76)
h ≈ 332.24 m
Therefore, the height of the cliff is approximately 332.24 meters.
Now let's calculate how far from the base of the cliff the diver hits the water. We can use the equation:
s = ut
Where:
s is the horizontal distance,
u is the horizontal velocity of the diver (2.0 m/s),
and t is the time it takes for the diver to reach the water (2.6 s).
Plugging in the values:
s = (2.0)(2.6)
s = 5.2 m
Therefore, the diver hit the water approximately 5.2 meters away from the base of the cliff.