Question

A pole vaulter uses the pole to lift him to a height of 7 m. What will his velocity be just before he hits the ground?

A) 137 m/s
B) 34.3 m/s
C) 11.7 m/s

Answer 1

h = height from which the pole vaulter falls = 7 m

v₀ = initial velocity of pole vaulter at height "h" = 0 m/s

v = final velocity of pole vaulter as he hits the ground = ?

g = acceleration due to gravity = 9.8 m/s²

m = mass of pole vaulter

using conservation of energy

kinetic energy just before he hits the ground = potential energy at the top + kinetic energy at the top

(0.5) m v² = (0.5) m v₀² + mgh

(0.5) v² = (0.5) v₀² + gh

v² = v₀² + 2gh

inserting the values

v² = (0)² + 2(9.8 x 7)

v = 11.7 m/s