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Carnegie Mellon University

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Problem 43 Hard Difficulty

A home run is hit in such a way that the baseball just clears a wall 21 $\mathrm{m}$ high, located 130 $\mathrm{m}$ from home plate. The ball is hit at an angle of $35^{\circ}$ to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (Assume the ball is hit at a height of 1.0 $\mathrm{m}$ above the ground.)

Answer

(a) $v _ { 0 } = 41.68 \mathrm { ms } ^ { - 1 }$
(b) $t = 3.8 \mathrm { s }$
(c) $v = 36.65 \mathrm { ms } ^ { - 1 }$

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Video Transcript

we can say first report A. The time for the ball to reach the fence would be t. This would be equaling Delta X, divided by the initial velocity in the ex direction. Of course, there isn't any acceleration in the ex direction, so we don't have to account for that term. This would be equaling 130 meters divided by the initial co sign of 35 degrees. And this is essentially equaling 159 meters divided by the initial and so we can set this time here. The bull must be 20 meters above its launch position and we're getting that simply from 21 meters minus one meter and this is giving us 20 meters Now. We know that Delta Y is going to be equaling the why initial t plus 1/2 times the acceleration in the UAE direction Times t squared and this is gonna give 20 meters equaling the initial this would be sign of 35 degrees. This would be multiplied by 159 meters, divided by the initial nine ISS, 4.90 meters per second squared multiplied by 159 meters divided by the initial quantity squared. Now the of the initial here is canceling out and essentially we find that the initial after solving is gonna be equaling 42 meters per second. This would be our final answer for part A for part B. We can then say that from part a time would be equaling 159 meters and this would be divided by 42 meters per second and this is giving us 3.8 seconds. So this would be our answer for part B. And then finally, for part C, we can say when the ball reaches the wall so this would be at T equaling 3.8 seconds. Ball reaches well. We can then say that V X velocity in the ex direction would be equaling. Of course, the initial velocity in the extraction and this would be 42 meters per second multiplied by co sign of 35 degrees. And this is giving us 34 meters per second. So this would be our velocity in the ex direction and meek and then say V, why final would be equaling b y initial, which we know to be 42 meters per second multiplied by sine of 35 degrees. And then it would be plus the acceleration in the why direction times t. So this would be a plus negative 9.80 meters per second squared multiplied by three 0.8 seconds and this is giving us negative 13 meters per second. So this would be downwards and then four part See. We can say that the magnitude of the final velocity would be equaling the square root of the X component squared, so this would be 34 meters per second quantity squared plus negative 13 meters per second quantity squared and we find that the magnitude of the final velocity is 36 meters per second. This would be our final answer for part C. That is the end of the solution. Thank you for watching

Carnegie Mellon University
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