Question

A man of mass 81 kg escapes from a burning building by jumping from a window situated 32 m above a catching net. The acceleration of gravity is 9.8 m/s2 . If air friction exerts a constant force of 103 N on him as he falls, what is his speed just before he hits the net?

Answer 1

The speed of man before he hits the ground is 23.35 m/s

Step-by-step explanation:

We know that:

Weight of Man - Force of Friction = Unbalanced Force

but, from Newton's 2nd Law of Motion:

unbalanced force = ma

Therefore,

W - F = ma

a = (W - F)/m

a = (mg - F)/m

where,

m = 81 kg

g = 9.8 m/s²

F = 103 N

a = [(81 kg)(9.8 m/s²) - 103 N]/81 kg

a = 8.52 m/s²

using 3rd equation of motion:

Vf² - Vi² = 2ah

here,

Vi = initial velocity = 0 m/s

Vf = Final Velocity before he hits ground = ?

Vf² - 0² = 2(8.52 m/s²)(32 m)

Vf = √545.28 m²/s²

Vf = 23.35 m/s