Question

An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish travels 7.9 m horizontally before hitting the water.

What is the velocity of the eagle? Round at the end of the calculation to the nearest tenth.

m/s

Answer 1

Vertical fall time t = sqrt(2h/g) = sqrt(2(6.1)/9.8) = sqrt(12.2/9.8) ≈ 1.1167 s.

Horizontal speed v = horizontal distance / t = 7.9 / 1.1167 ≈ 7.0799 m/s ≈ 7.1 m/s (to the nearest tenth).