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Carnegie Mellon University

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Problem 11 Medium Difficulty

A placekicker must kick a football from a point 36.0 $\mathrm{m}$ (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 $\mathrm{m}$ high. When kicked, the ball leaves the ground with a speed of 20.0 $\mathrm{m} / \mathrm{s}$ at an angle of $53.0^{\circ}$ to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising or while falling?

Answer

a. the ball clears the crossbar by 0.89$\mathrm { m }$
b. falling

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Video Transcript

over per day. We can first say the time required for the ball to travel. We can Delta X equaling 36.0 meters. Ah, would be this would be equaling v ex initial times t and this would be equaling v initial where we can say to say, visa vie a co sign of Fada multiplied by t Or we can say that then t would be equaling. Delta acts over the initial co sign of data. Now that vertical displacement of this far, we can say Delta y would be equaling the why initial t plus 1/2 ace of Y t squared. And so the vertical distance by which the ball clears the ball at the bar is D. This would be Delta y minus h. So again, vertical distance, um by which oh clears four. And so we can then say that D continue on for per day. D equals the initial sign of 53 degrees multiplied by 36.0 meters, divided by the initial co sign of 53 degrees. Here, of course, the initial cancels out plus 1/2 multiplied by negative 9.80 meters per second squared multiplied by 36.0 meters, divided by this would be 20.0 meters per second multiplied by co sign of 53 degrees quantity squared and then this would be minus 3.5 meters and so D is then equaling 0.89 meters. The fact that this is positive means that the ball the ball clears the crossbars by 0.89 meters. This would be our full answer again. This would be minus 3.5 meters. I apologize that it's crowded and so for part B, we know the the ball reaches the plane of the goal post. AT T equaling Delta acts over the initial co sign. If Ada so now we can solve. This would be 36.0 meters, divided by 20.0 meters per second times co sign of 53 degrees and this is equaling 2.99 seconds and so at this time the vertical velocity V Y final equaling V Y initial plus acceleration War Direction t uh, this would be equaling 20.0 meters per second time sign of 53 degrees plus negative 9.80 meters per second squared multiplied by 2.99 seconds. And we find that V why final is equaling negative 13.3 meters per second. The fact that this is negative the fact that this is less than zero means that the ball passes rather better yet, we can say when the ball crosses the crowbar for the crossbar. Uh, the ball has passed. It's peak and is now descending. So this would be our final answer for part B. That is the end of the solution. Thank you for watching.

Carnegie Mellon University
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