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A baseball is thrown horizontally off a cliff with a velocity of 15.9 m/s. It lands 35.9 meters away. Assuming that gravity is 9.8 m/s\(^2\), how high (dy) is the cliff?

Answer :

The height of the cliff, given that the baseball landed 35.9 m away is 25.03 m

How to determine the time

First, we shall determine the time taken for the baseball to land. This can be obtained as follow:

  • Horizontal velocity (u) = 15.9 m/s
  • Horizontal distance (s) = 35.9 m
  • Time (t) =?

s = ut

35.9 = 15.9 × t

Divide both sides by 15.9

t = 35.9 / 15.9

t = 2.26 s

How to determine the height of the cliff

We can obtain the height of the cliff as follow:

  • Acceleration due to gravity (g) = 9.8 m/s²
  • Time (t) = 2.26 s
  • Height (h) =?

h = ½gt²

h = ½ × 9.8 × 2.26²

h = ½ × 9.8 × 5.1076

h = 25.03 m

Thus, we can conclude that the cliff is 25.03 m high

Learn more about motion under gravity:

https://brainly.com/question/22719691

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