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A heavy object is dropped from a vertical height of 9.0 m. What is its speed when it hits the ground?

1 Answer

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Answer:

13.42 m/s

Step-by-step explanation:

To find the speed when it hits the ground, we will use the following equation


v_f^2=v_i^2+2a(\Delta y)

Where vf is the final velocity, vi is the initial velocity, a is the acceleration due to gravity, and Δy is the height.

Replacing vi = 0 m/s, a = 10 m/s², and Δy = 9.0 m, we get:


\begin{gathered} v_f^2=0^2+2(10)(9) \\ v_f^2=180 \\ v_f=√(180) \\ v_f=13.42\text{ m/s} \end{gathered}

Therefore, the speed when it hits the ground is 13.42 m/s

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