A sky diver of mass 68 kg jumps from a
slowly moving aircraft and eventually reaches her terminal velocity 46m/s
Now consider the same sky diver jumping out of a faster airplane. At some point before she reaches her terminal velocity, she has velocity v2 of magnitude 46.5 m/s in the direction 30� below horizontal.
What is the magnitude of the sky diver’s acceleration at that point in time?
Answer in units of m/s^2
use 9.8 for acceleration of gravity.
To find the magnitude of the sky diver's acceleration at the given point in time, we need to break down the velocity vector into its horizontal and vertical components.
First, let's find the vertical component of the velocity (v2y) by using the given angle of 30 degrees below the horizontal. We can use trigonometry to determine this value.
v2y = v2 * sin(30)
= 46.5 * sin(30)
≈ 23.25 m/s
Next, we can find the net force acting on the sky diver when the velocity is v2. At this point, the sky diver has not yet reached her terminal velocity, so there is a net force acting on her.
The net force is given by the equation:
Fnet = mass * acceleration
Since the only force acting on the sky diver during the free fall is gravity, we can write:
Fnet = mass * acceleration = mass * g
Where g is the acceleration due to gravity.
So, we can solve for the acceleration:
acceleration = Fnet / mass = g
Given that g = 9.8 m/s^2, the magnitude of the sky diver's acceleration at that point is 9.8 m/s^2.