A relative of yours belly flops from a height of 2.50 m (ouch!) and stops moving after descending 0.500 m underwater. Her mass is 62.5 kg.
What was the magnitude of her acceleration in the pool? Assume that it is constant.
How long was she in the water before she stopped moving?
What was the magnitude of the average net force exerted on her after she hit the water until she stopped?
To find the magnitude of her acceleration in the pool, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where vf is the final velocity (0 m/s since she stopped moving), vi is the initial velocity (0 m/s since she started from rest), a is the acceleration, and d is the displacement.
Since she stopped moving, vf is 0 m/s. The displacement, d, is the height of the fall minus the distance underwater:
d = 2.50 m - 0.500 m = 2.00 m
Plugging in the known values into the equation:
0^2 = 0^2 + 2a(2.00 m)
0 = 4a
Therefore, the magnitude of her acceleration in the pool is 0 m/s^2.
To find how long she was in the water before stopping, we can use another kinematic equation:
vf = vi + at
Since she stopped moving, vf is 0 m/s. The initial velocity, vi, is also 0 m/s. The acceleration, a, is 0 m/s^2. Let's solve for t:
0 = 0 + 0 * t
0 = 0
This means that she stopped instantly upon hitting the water, so the time she was in the water before stopping is 0 seconds.
To determine the magnitude of the average net force exerted on her from hitting the water until she stopped, we can use Newton's second law of motion:
Fnet = ma
The mass, m, is given as 62.5 kg. The acceleration, a, is 0 m/s^2. Let's calculate the force:
Fnet = 62.5 kg * 0 m/s^2
Fnet = 0 N
Therefore, the magnitude of the average net force exerted on her after hitting the water until she stopped is 0 N.