Answer:
v = 36.05 m/s
Explanation:
It is given that,
Initial speed of the parachutist, u = 14.8 m/s
Altitude, h = 55.1 m
The acceleration due to gravity, [tex]a=9.81\ m/s^2[/tex]
Let v is the speed of the shoe just before it hits the ground. The third equation of motion is given by :
[tex]v^2=u^2+2ah[/tex]
[tex]v^2=(14.8)^2+2\times 9.81\times 55.1[/tex]
v = 36.05 m/s
So, the speed of the show just before it hits the ground is 36.05 m/s. Hence, this is the required solution.
