An animal-rescue plane flying due east at 33.0 m/s drops a bale of hay from an altitude of 58.0 m. If the bale of hay weighs 173 N, what is the momentum of the bale the moment it strikes the ground?
21=tff
M*g = 173.
M = 173/g = 173/9.8 = 17.7kg = Mass of the hay.
V^2 = Vo^2 + 2g*h = 0 + 19.6*58 = 1137,
V = 33.7 m/s.
Momentum = M*V =
To calculate the momentum of the bale of hay when it strikes the ground, we need to find its velocity just before impact.
We know that the plane is flying due east at a speed of 33.0 m/s, which means the horizontal component of the bale's velocity is also 33.0 m/s. Since the bale is dropped vertically, it only has a vertical velocity component.
The vertical distance it falls is 58.0 m, and we can use the kinematic equation that relates displacement, initial velocity, time, and acceleration to find the time it takes for the bale to fall:
Δy = V yi × t + 0.5 × g × t^2
Here, Δy represents the vertical displacement, V yi is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
Since the bale is dropped, its initial vertical velocity is 0 m/s, so the equation simplifies to:
58.0 m = 0 × t + 0.5 × 9.8 m/s^2 × t^2
Rearranging the equation:
4.9 t^2 = 58.0
Dividing both sides by 4.9:
t^2 = 11.8
Taking the square root of both sides:
t ≈ 3.43 s
Now, we can find the final vertical velocity of the bale just before it hits the ground using the equation:
V yf = V yi + g × t
V yf = 0 + 9.8 m/s^2 × 3.43 s
V yf ≈ 33.71 m/s (rounded to two decimal places)
Therefore, the velocity of the bale just before it hits the ground is approximately 33.71 m/s. To find the momentum, we multiply the velocity by the mass:
Momentum = 33.71 m/s × 173 N
Momentum ≈ 5,828.63 kg⋅m/s (rounded to two decimal places)
Thus, the momentum of the bale just before it strikes the ground is approximately 5,828.63 kg⋅m/s.