A)a 72 kg man jumps from a window 1.2 m above a sidewalk. the acceleration of gravity is 9.81 m/s2. what is his speed just before his feet strike the pavement? answer in units of m/s
B) if the man jumps with his knees and ankles locked, the only cushion for his fall is approximately .48 cm in the pads of his feet. calculate the magnitude of the average force exerted on him by the ground in this situation. answer in units of N.
To find the speed of the man just before his feet strike the pavement, we can use the principle of conservation of energy. We can assume that the initial potential energy when the man is at the window gets converted into kinetic energy just before he hits the ground. We can use the following equation:
Potential Energy + Kinetic Energy = Total Energy
1/2 * mass * velocity^2 + mass * gravity * height = 1/2 * mass * speed^2
Let's now plug in the known values and solve for speed:
Given:
Mass (m) = 72 kg
Gravity (g) = 9.81 m/s^2
Height (h) = 1.2 m
Potential Energy = m * g * h
Potential Energy = 72 kg * 9.81 m/s^2 * 1.2 m = 849.1 Joules
Using the equation, we can solve for speed:
1/2 * (72 kg) * velocity^2 + (72 kg) * (9.81 m/s^2) * 1.2 m = 849.1 J
36 * velocity^2 + 847.68 = 849.1
36 * velocity^2 = 849.1 - 847.68
36 * velocity^2 = 1.42
velocity^2 = 1.42 / 36
velocity^2 = 0.0394
velocity ≈ √0.0394
velocity ≈ 0.1982 m/s
Therefore, the speed just before the man's feet strike the pavement is approximately 0.1982 m/s.
Moving on to part B, to calculate the magnitude of the average force exerted on the man by the ground, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In this situation, the acceleration will be due to deceleration, caused by the pads of his feet cushioning his fall.
Since the man jumps with his knees and ankles locked, assuming a uniform compression of the pads during landing, we can calculate the average acceleration (a) using the following equation:
a = (vf - vi) / Δt
Where:
vf = final velocity (0 m/s since he stops)
vi = initial velocity (0.1982 m/s as calculated in part A)
Δt = time taken to decelerate (unknown)
Now, we need to find the time required to decelerate, which can be calculated by considering the distance the pads compress. Given that the cushion for his fall is approximately 0.48 cm in the pads of his feet, we convert it to meters:
Compression distance (d) = 0.48 cm = 0.48 / 100 m
Since the average force exerted on the man by the ground is equal to the mass multiplied by the average acceleration, we can now proceed with the calculation.
Average Force = mass * average acceleration
Let's substitute the known values:
mass (m) = 72 kg
average acceleration (a) = (vf - vi) / Δt
From vi = 0.1982 m/s and vf = 0 m/s, we can simplify:
a = (0 - 0.1982) / Δt
a = -0.1982 / Δt
Now, we need to find time (Δt) by rearranging the equation for distance:
d = vi * Δt + (1/2) * a * (Δt^2)
Substituting the known values:
0.48 / 100 m = 0.1982 m/s * Δt + (1/2) * (-0.1982 / Δt) * (Δt^2)
0.0048 m = 0.1982 Δt - 0.0991 Δt
0.0048 = 0.0991 Δt
Δt = 0.0048 / 0.0991
Δt ≈ 0.0486 seconds
Now, let's calculate the average acceleration:
a = -0.1982 / 0.0486
a ≈ -4.082 m/s^2
Finally, we can calculate the average force using the equation:
Average Force = mass * average acceleration
Average Force = 72 kg * -4.082 m/s^2
Average Force ≈ -294.8 N
Therefore, the magnitude of the average force exerted on the man by the ground in this situation is approximately 294.8 N.
A) V = sqrt(2 g H)
B) (1/2)MV^2 = Force*("cushion distance")