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: **

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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The bikers successfully landed on the very edge of the other side, what was his speed on the runway? The default is: α = 53 °, d = 40 m, h1 = 100 m, h2 = 15 m.