Question

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?

11.0 m

260 m

35.8 m

848 m

Answer 1

Answer:

  35.8 m

Explanation:

An appropriate formula is ...

  $V_f^2-V_i^2=2ad$

The initial velocity is 0, and the acceleration due to gravity is 9.8 m/s², so the distance, d, is ...

  $d=\dfrac{26.5^2}{2\cdot 9.8}\approx 35.8 \quad\text{m/s}$