# An animal-rescue plane due east at 50 meters per second drops a bale of hay from an altitude of...

## Question:

An animal-rescue plane due east at 50 meters per second drops a bale of hay from an altitude of 62 meters. The acceleration due to gravity is 9.81 meters per second squared. If the bale of hay weighs 191 newtons what is the momentum of the bale the moment it strikes the ground? Answer in units of kilograms times meters per second.

## Freefall:

When a rescue plane flying horizontally drops a bale of hay, the bale inherits the horizontal speed of the rescue plane; the vertical velocity increases due to gravity. The momentum on impact equals the product of the mass and the velocity on impact.

If the weight of the bay is {eq}W = 191\ \text{N} {/eq} on earth's gravitational field of acceleration {eq}g = 9.81\ \text{m/s}^2 {/eq}, the mass m is the weight divided by the acceleration, i.e. {eq}m = \frac{W}{g} {/eq}

Considering the downward motion, we have the following, taking the downward direction as positive:

• The initial velocity is {eq}u=0 {/eq}
• The vertical distance is {eq}s = 50\ \text{m} {/eq}
• the downward acceleration is {eq}a = g= 9.81\ \text{m/s}^2 {/eq}

Using a suitable equation, the final velocity v is:

{eq}\begin{align*} v^2 &= u^2+2gs\\ v &= \sqrt{u^2+2gs}\\ &= \sqrt{0^2+2(9.81\ \text{m/s}^2)(50\ \text{m})} &\approx 31.3\ \text{m/s} \end{align*} {/eq}

The speed v during the landing is the Pyhtagorean hypotenuse of the horizontal speed {eq}v_x = 50\ \text{m/s} {/eq} and vertical speed {eq}v_x =77.8\ \text{m/s} {/eq} i.e.

{eq}\begin{align*} v &= \sqrt{(31.3\ \text{m/s})^2+(50\ \text{m/s})^2}\\ &\approx 59.0\ \text{m/s} \end{align*} {/eq}

The momentum {eq}\rho {/eq}, equals the product of the mass m and the speed v i.e.

{eq}\begin{align*} \rho &= mv\\ &= \frac{Wv}{g}\\ &= \frac{191\ \text{N}(59.0\ \text{m/s})}{9.81\ \text{m/s}^2}\\ &\color{blue}{\approx \boxed{1\ 149\ \text{kg.m/s}}} \end{align*} {/eq} 