Question

A parachutist's fall to Earth is determined by two opposing forces. A gravitational force of 605 N acts on the parachutist. After 3 s, she opens her parachute and experiences an air resistance of 689 N. At what speed (in m/s) is the parachutist falling after 10 s?

Answer 1

Answer:15.8 m/s

Explanation:

Given

Gravitational Force $F_g=605 N$

Air resistance $F_d=689 N$

mass of Parachutist $=\frac{605}{9.8}=61.73 kg$

Velocity after $3 s$

$v=u+at$

$v=0+9.8\times 3$

$v=29.4 m/s$

Now Parachutist opens the parachute

Net for on Parachutist

$605-689=-84$

$84 N$ drag force

therefore deceleration $\frac{84}{61.73}=1.36 m/s^2$

velocity after 10 s is

$v_1=v+at$

$v_1=29.4-1.36\times 10$

$v_1=15.8 m/s$