A 52.2 kg parachutist lands moving straight downward with a speed of 3.75 m/s. If the parachutist comes to rest with constant acceleration over a distance of 0.794 m, what force does the ground exert on her?
average speed during crash = 3.75/2
so
time to stop = .794 / (3.75/2)
acceleration = change in speed / change in time
or
a = (0 - -3.75) / [.794 / (3.75/2) ]
and
F = m a = 52.2 a
To find the force exerted by the ground on the parachutist, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
Given:
Mass of the parachutist, m = 52.2 kg
Initial velocity of the parachutist, u = 3.75 m/s (moving downward)
Distance traveled by the parachutist, s = 0.794 m
Final velocity of the parachutist, v = 0 m/s (comes to rest)
First, we need to find the acceleration of the parachutist using the equation:
v^2 = u^2 + 2as
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0^2 - 3.75^2) / (2 * 0.794)
a = -14.14 m/s^2 (negative because the parachutist decelerates)
Now, we can use Newton's second law to find the force:
F = ma
Substituting the mass and acceleration:
F = 52.2 kg * (-14.14 m/s^2)
F = -737.628 N
The negative sign indicates that the force exerted by the ground is in the upward direction (opposite to the motion). So, the force exerted by the ground on the parachutist is approximately 737.628 N upwards.