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Carnegie Mellon University



Problem 42 Hard Difficulty

A ball is thrown straight upward and returns to the thrower's hand after 3.00 s in the air. A second ball thrown at an angle of $30.0^{\circ}$ with the horizontal reaches the same maximum height as the first ball. (a) At what speed was the first ball thrown? (b) At what speed was the second ball thrown?


(a) $v _ { \mathrm { by } } = 14.7 \mathrm { ms } ^ { - 1 }$
(b) $v _ { 0 } = 29.4 \mathrm { ms } ^ { - 1 }$


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Video Transcript

so we could first calculate the speed of the first ball we can save E Y. Final equals V. Why initial plus g t. And in this case, T is going to be equaling the time it takes to reach the maximum height. It takes three seconds for the ball to come to go straight upwards, reach its maximum height. Elin, come right back down. And it's under a constant acceleration throughout that entire time. So the time it takes to reach the highest point is gonna be half of three seconds. So this would be 3.0 seconds divided by two. This would be 1.50 seconds. And we can then say that the initial velocity with the final velocity equaling zero, the initial velocity would be equaling negative g t. And so this would be equally negative. Negative 9.80 meters per second squared multiplied by 1.50 seconds. And this is equaling 14.7 meters per second. And this would be, of course, upwards. So this would be the initial speed of the first ball. This would be your answer for part A For part B. We want to see what speed was the second ball thrown well. First, we should try to find We can actually say that the second ball we know to be thrown at an angle of 14 of Ah rather an angle of 30 degrees. And so we simply have that. The why initial must equal 14.7 meters per second in order to reach the same height that the first ball did. However, this is gonna be equal in the initial velocity times sign of 30 degrees. And so we want to find that initial velocity that initial velocity would be 14.7 meters per second, divided by sign of 30 degrees, and this is equaling 29.4 meters per second so that so the second ball needs to be thrown with a initial velocity of 29.4 meters per second if it's being thrown 30 degrees from the horizontal in order to reach the same heights that the first ball did, given that it had a, uh, the same initial velocity in the UAE direction 14.7 meters per second. So this would be the initial velocity of the second ball thrown at an angle 30 degrees above the horizontal. That is the end of the solution. Thank you for watching

Carnegie Mellon University
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