Question

A 120 kg base jumper stands on a bridge at an elevation of 250 meters above the water. To the nearest tenth, what is her speed just before she strikes the water

Answer 1

The answer to your question is vf = 70 m/s

Step-by-step explanation:

This is a free fall problem and in these kinds of problems, the mass is not important to calculate the speed.

Data

mass = 120 kg

height = 250 m

g = 9.81 m/s²

Formula

h =
`(vf^(2))/(2g)`

Solve for vf

vf² = 2gh

vf =
`√(2gh)`

Substitution

vf =
`√(2(9.81)(250))`

Simplification

vf =
`√(4905)`

Result

vf = 70 m/s