The Cattle and Hay Problem

An airplane is dropping bales of hay to cattle. The pilot releases the bales at 150m above the level ground when the plane if flying at 75m/s in a direction above \(55^\circ\) above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales land at the point where the cattle is stranded?


Figure 1

The vertical part of the motion is projectile motion but the horizontal part is not.
The hay falls 150m with a downward acceleration g. It must travel a horizontal distance with constant horizontal velocity.
The bale has inital velocity components $$ v_x=v_0cos\alpha=75cos55^\circ=43m/s$$
The vertical component is
$$ v_y=v_0sin\alpha=75sin55^\circ=61.4m/s$$
Now, \(y_0=150m\)
For vertical motion we use the equation $$ y-y_0=v_yt+\frac{1}{2}a_yt^2$$ Putting the appropriate values of \(y\), \(v_y\) and \(a_y\), we have$$ -150=6.14t-4.9t^2$$
. On solving the above quadratic formula for t, we get $$ t=6.27\pm8.36$$.
Since t cannot be negative we take the positive value which is \(t=14.6s\)
Then putting the value of t in the equation for horizontal motion we have
$$ x=v_xt+\frac{1}{2}a_xt^2$$
Now, \(a_x=0\), so the second term in the equation becomes zero.
Hence $$x=(43)(14.6)=630m$$
So, the airplane needs to release the bales \(630m\) in front of the cattle.

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