An animal rescue plane flying due east at 36.0 m/s...?

Question

An animal rescue plane flying due east at 36.0 m/s drops a bale of hay from an altitude of 60.0 m. If the bale of hay weighs 175 N, what is the momentum of the bale the moment it strikes the ground?

For some reason, in addition to the momentum the teacher gave me the angle at which it was below the horizon. Could someone please show me the work in detail to this problem? Please explain what all of your variables mean, just in case my text uses different variables.

Thanks!

Answer 1

projectile motion
i, j are unit vectors
in the x-direction
assume the bail of hay was ejected from the plane versus the horizontal at angle @ to the vertical or angle 90-@ to the horizon

x=0
vx=36
ax=0

in the y-direction
y=60
vy=0
ay=9.8=g

y=1/2gt^2

the time it takes the hay to hit the ground is sqrt(2*60/9.8)
=3.5 s

it will travel down the range 36*3.5=126 m

the angle it is ejected=arctan(y/x)
=arctan(60/126)=25.5 degrees

assuming no air resistance it hits the ground at 9.8*3.5=34.3 m/s in the y direction and 36 m/s in the x direction. the magnitude of both speeds
is sqrt(34.3^2+36^2)=49.7 m/s

the momentum=mv=(175/9.8)*49.7=888 kg-m/s

Answer 2

1) Coordinate system
Let Y be up and X horizontal and the origin on the surface of the Earth.

2) Initial state
The hay is released @ X = 0 and Y = H with horizontal speed equal to V. The mass of the hay is m

3) Equations of motion - X direction
d[dX/dt]/dt = 0; dX/dt = V; X = Vt

4) Equations of motion - Y direction
d[dY/dt]/dt = -g; dY/dt = -gt; Y = H - g(t^2) / 2

5) Let T be the time of impact
H - g(T^2) / 2 = 0;
T^2 = 2H / g

6) Let Q be the magnitude of the momentum @ impact
Q = m * S; where S is the speed @ T and m is the mass of the hay
Q = m * SQRT( (dX/dt)^2 + (dY/dt)^2 )
= m * SQRT( V^2 +2gH)
= m * V * SQRT( 1 + 2gH/(V^2))