a physics student dives horizontally with an initial velocity of 2.6 m/s off the Edge off a cliff, 120 meters high above the water surface. how far from the base of the cliff will the student land on the Water's surface?
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a physics student dives horizontally with an initial velocity of 2.6 m/s off the Edge off a cliff, 120 meters high above the water surface. how far from the base of the cliff will the student land on the Water's surface?
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Answer:
We can solve this problem using the equations of motion for a projectile. First, we need to break down the initial horizontal and vertical velocities of the physics student:
Initial horizontal velocity (Vx) = 2.6 m/s (given in the question)
Initial vertical velocity (Vy) = 0 m/s (the student dives horizontally)
Next, we need to determine the time it takes for the student to fall from the cliff to the water surface. We can use the following kinematic equation to find the time (t):
h = (1/2)gt^2
where h is the height of the cliff (120 m), g is the acceleration due to gravity (9.81 m/s^2), and t is the time.
Rearranging the equation, we get:
t = sqrt((2h)/g) = sqrt((2 x 120)/9.81) = 5.11 s
Therefore, it takes 5.11 seconds for the student to fall from the cliff to the water surface.
Now, we can use the following equation to find the horizontal displacement (d):
d = Vx x t
Substituting the values, we get:
d = 2.6 m/s x 5.11 s = 13.30 m
Therefore, the physics student would land on the water's surface 13.30 meters away from the base of the cliff.