High School

An object is dropped from the edge of a cliff and is moving at 26.5 m/s just before it hits the ground. How high is the cliff?

A. 848 m
B. 11.0 m
C. 35.8 m
D. 260 m

Answer :

Final answer:

The height of the cliff is 35.8 meters.

Explanation:

The height of the cliff can be determined by using the equations of motion. Since the object is dropped and its initial velocity is zero, we can use the equation

[tex]h = (1/2)gt^2,[/tex]

where h is the height of the cliff, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken to hit the ground.

Substituting the given values (initial velocity = 0, final velocity = 26.5 m/s) into the equation for the time of flight of an object dropped from a height gives

[tex]h = (1/2)gt^2 = (1/2)(9.8 m/s^2)(2.711s)^2[/tex]

t = sqrt((2h)/g) = sqrt((2h)/9.8)

= 2.711s.

Solving for h gives us

= 35.8 m.

Therefore, the height of the cliff is 35.8 meters.

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