A parachutist descending at a speed of 19.6
m/s loses a shoe at an altitude of 41.6 m.
What is the velocity of the shoe just before
it hits the ground? The acceleration of gravity
is 9.81 m/s2.
Answer in units of m/s
Vf^2 = Vo^2 + 2gd,
Vf^2 = (19.6)^2 + 2*9.8*41.6 = 1199.52,
Vf = 34.6m/s.
To find the velocity of the shoe just before it hits the ground, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
where
vf = final velocity (what we are looking for)
vi = initial velocity (given as 19.6 m/s)
a = acceleration (gravity, -9.81 m/s^2 since it acts in the opposite direction)
d = distance traveled (41.6 m)
Plugging the values into the equation, we have:
vf^2 = (19.6 m/s)^2 + 2(-9.81 m/s^2)(41.6 m)
Calculating it:
vf^2 = 384.16 m^2/s^2 + (-812.0352 m^2/s^2)
vf^2 = -427.8752 m^2/s^2
Since the velocity cannot be negative, we take the square root of the positive value:
vf = √(427.8752 m^2/s^2)
Calculating it:
vf ≈ 20.68 m/s
So, the velocity of the shoe just before it hits the ground is approximately 20.68 m/s.
To find the velocity of the shoe just before it hits the ground, we can use the principle of conservation of energy.
The initial potential energy of the shoe at an altitude of 41.6 m is given by the equation:
Potential energy = mass * gravity * height
Since the shoe is dropped from rest, its initial kinetic energy is zero.
The final kinetic energy of the shoe just before it hits the ground is given by the equation:
Kinetic energy = (1/2) * mass * velocity^2
By equating the initial potential energy to the final kinetic energy, we can solve for the velocity.
Potential energy = Kinetic energy
mass * gravity * height = (1/2) * mass * velocity^2
Dividing both sides of the equation by mass and canceling the mass term:
gravity * height = (1/2) * velocity^2
Simplifying further:
velocity^2 = 2 * gravity * height
Taking the square root of both sides of the equation, we can find the velocity:
velocity = sqrt(2 * gravity * height)
Plugging in the values for gravity (9.81 m/s^2) and height (41.6 m) into the equation:
velocity = sqrt(2 * 9.81 * 41.6)
Calculating this expression:
velocity = 28.0 m/s
Therefore, the velocity of the shoe just before it hits the ground is 28.0 m/s.