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Carnegie Mellon University

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Problem 57 Hard Difficulty

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 25.0 $\mathrm{m} / \mathrm{s}$ . The first is thrown at an angle of $70.0^{\circ}$ with respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first for both to arrive at the same time?

Answer

(a) $\theta _ { 2 } = 20 ^ { \circ }$
(b) $t = 3.05 \mathrm { s }$

Discussion

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CB

Calanao, B.

November 22, 2020

CB

Calanao, B.

November 22, 2020

Video Transcript

over party. We can first start out by saying doubts. A wife equaling V Y initial t plus 1/2 times the acceleration of the Y direction Times t squared. Now we can apply this the first snow ball first and we can say that this would be essentially zero equaling 25 meters per second sign of 70 degrees multiplied by tea. And this would be plus 1/2 most supplied by negative 9.80 meters per second squared multiplied by t squared. Of course, this tea will cancel out with one of these teas and t is gonna be equaling two times 25 meters per second, multiplied by sine of 70 degrees. And this would be all divided by 9.80 meters per second squared. And this is giving us 4.79 seconds now. The horizontal displacement during this time would be equaling V X initial t, and this would be equaling 25 meters per second co sign of 70 degrees multiplied by 4.79 seconds. And this is giving us 41.0 meters now. At this point, we can say that Apply all of these to the second snowball. So for the second snowball, we have then zero equaling again 25 meters per second. Now we don't know the angle of launch, so this will simply be signed off. Beta multiplied by t some too, plus 1/2 times negative, 9.80 meters per second square multiplied by T sub to quantity squared and we find that gun t's up to is gonna be equaling two times 25 meters per second, multiplied by sine of data divided by 9.8 meters per second squared. And this is giving us essentially 5.10 seconds multiplied by sine of theta. Whatever that launch angle is now, we require this the first, the horizontal range of the first snowball to be equal to the horizontal range of the second snowball. And so we can say doubt tow acts is equaling the ex initial t's up to and so this would be 25 meters per second multiplied by co sign of Fada multiplied by 5.10 seconds multiplied by sine of fada. So this would simply be equaling 41.0 meters because we know that from the previous we had, we achieved a horizontal range of 41.0 meters. Now, essentially, we can say that then sign a true economic identity. That's pretty useful sign of state a coast. And if Ada is equaling sign of tooth data and so we can then say that essentially sign of Tooth Ada is gonna be equaling 41.0 meters, divided by 25 meters per second, multiplied by 5.10 seconds. And this is giving us 0.321 so quickly we can say that fate a would be equaling 1/2 park sign of 0.321 and moving over here, we find that Seita is gonna be equaling approximately 20.0 degrees. This would be our final answer for the angle of launch for that second snowball. Now, for part B, we can say that the time of flight for the first no ball tifs of one is equaling again. 4.79 seconds. And for the second snowball, this would be equaling 5.10 seconds time sign of 20.0 degrees. And so this is equaling 1.7 four seconds. So this would be the time of flight for the second snowball if they were to arrive simultaneous if they were to arrive simultaneously, then the time delay between the first and second snowballs is very easy to calculate. Delta T would simply be equaling t someone minus T's up to. So this would be 4.79 seconds minus 1.74 seconds, and we find that then Delta T is going to be equaling 3.5 seconds. This would be our final answer for Part B. That is the end of the solution. Thank you for watching.

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