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An errand of mercy. An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150 $\mathrm{m}$ above the level ground when the plane is flying at 75 $\mathrm{m} / \mathrm{s} 55^{\circ}$ above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales will land at the point where the cattle are stranded?

630 m

Physics 101 Mechanics

Chapter 3

Motion in a Plane

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Cornell University

University of Michigan - Ann Arbor

University of Washington

Simon Fraser University

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So here Delta, why is equaling negative 150 meters? The projectile is falling Ah, falling from a height of 150 meters. We know that the initial equals 75 meters per second and we know that the angle of launch equals 55 degrees above the horizontal. We can say that Delta y equals B Y initial t plus 1/2 g t squared and this becomes negative. 150 equaling 75. Sign of 55 degrees times T minus 4.9 t squared. This is a quadratic equation, so you can simply use your teeth. I 84 85 89 to solve for this and you can say t equals you. Confined that t equals 14.63 seconds. So to find the total horizontal displacement, we have the ex initial T plus 1/2 A T squared. There wasn't any acceleration in the extraction during free fall. And so we can say that the total horizontal displacement is going to be 75 co sign of 55 degrees times time of 14.63 and we're getting that the total horizontal displacement is going to be equal to approximately 629 meters. That is the end of the solution. Thank you for watching.

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