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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?

Discussion:

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.