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A stone is thrown horizontally with a speed of 15 m/s from the top of a vertical cliff at the edge of a lake. If the stone hits the water 2.0 s later, the height of the cliff is closest to:________

a. 90.0 m
b. 80.0 m
c. 100 m
d. 110 m

User Drojf
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1 Answer

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Final answer:

The height of the cliff can be calculated using the equations of motion. By applying the equation h = (1/2)gt^2, we can solve for the height of the cliff as 19.6 m. Therefore, the closest option is c - 100 m.

Step-by-step explanation:

The height of the cliff can be calculated using the equations of motion. Since the stone is thrown horizontally, its initial vertical velocity is 0 m/s. The only force acting on the stone in the vertical direction is gravity (9.8 m/s^2), which causes it to accelerate downward. We can use the equation h = (1/2)gt^2 to relate the height of the cliff (h) to the time it takes for the stone to hit the water (t).

Plugging in the values:

  • g = 9.8 m/s^2 (acceleration due to gravity)
  • t = 2.0 s (time taken to hit the water)

We can rearrange the equation to solve for h: h = (1/2)gt^2.

Substituting the given values, we get h = (1/2)(9.8 m/s^2)(2.0 s)^2 = 19.6 m.

Therefore, the height of the cliff is closest to 19.6 m, which corresponds to option c.

User Rodolfo Carvalho
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