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.