A person jumps from the roof of a house 4.9 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. The mass of his torso (excluding legs) is 46 kg.
(a) Find his velocity just before his feet strike the ground.
1 m/s
(b) Find the average force exerted on his torso by his legs during deceleration.
Magnitude
To find the velocity just before the person's feet strike the ground, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (0 m/s as the person jumps vertically downwards)
a = acceleration due to gravity (-9.8 m/s^2, assuming no air resistance)
s = displacement (4.9 m)
Rearranging the equation, we get:
v^2 = 0^2 + 2(-9.8)(4.9)
v^2 = -2(9.8)(4.9)
v^2 = -96.04
Since velocity cannot be negative in this context, we take the positive square root:
v = √96.04
v ≈ 9.8 m/s
Therefore, the velocity just before the person's feet strike the ground is approximately 9.8 m/s.
To find the average force exerted on their torso by their legs during deceleration, we can use Newton's second law of motion:
F = m * a
Where:
F = force (unknown)
m = mass of the torso (46 kg)
a = acceleration (unknown)
We can find the acceleration using the equation:
a = (vf - vi) / t
Where:
vf = final velocity (0 m/s, as the person's torso comes to rest)
vi = initial velocity (9.8 m/s)
t = time taken to decelerate (unknown)
Rearranging the equation, we get:
t = (vf - vi) / a
t = (0 - 9.8) / a
t = -9.8 / a
We can substitute this value of t into the equation for force:
F = m * (-9.8 / a)
The acceleration can be found by using the displacement and the initial velocity:
s = (vi^2 - vf^2) / (2 * a)
0.70 = (9.8^2 - 0) / (2 * a)
0.70 * (2 * a) = 96.04
1.4a = 96.04
a = 96.04 / 1.4
a ≈ 68.6 m/s^2
Now we can substitute this value of a into the equation for force:
F = 46 * (-9.8 / 68.6)
F ≈ -6.612 N
Since force cannot be negative in this context, we take the magnitude:
|F| ≈ 6.612 N
Therefore, the average force exerted on the person's torso by their legs during deceleration is approximately 6.612 N.