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Carnegie Mellon University

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Problem 40 Hard Difficulty

A farm truck travels due east with a constant speed of 9.50 $\mathrm{m} / \mathrm{s}$ along a horizontal road. A boy riding in the back of the truck tosses a can of soda upward (Fig. P3.40) and catches it at the same location in the truck bed, but 16.0 $\mathrm{m}$ farther down the road. Ignore any effects of air resistance, (a) At what angle to the vertical does the boy throw the can, relative to the moving truck? (b) What is the can's initial speed relative to the truck? (c) What is the shape of the can's trajectory as seen by the boy? (d) What is the shape of the can's trajectory as seen by a stationary observer on the ground? (e) What is the initial velocity of the can, relative to the stationary observer?

Answer

(a) $\theta = 0 ^ { \circ }$
(b) $u _ { 0 } = 8.23 \mathrm { ms } ^ { - 1 }$
(c) The shape of the can's trajectory ns seen by the boy in the truck is $n$
straight line.
(d) Parabola
(e) $\vec { t } _ { 0 } = 9.5 \mathrm { m } \mathrm { s } ^ { - 1 } \mathrm { i } + 8.23 \mathrm { m } \mathrm { s } ^ { - 1 } \mathrm { j }$ and $\left| \vec { v } _ { 0 } \right| = 12.57 \mathrm { m } \mathrm { s } ^ { - 1 }$

Discussion

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Video Transcript

the answer. Any questions? We should first find the time of flight. And so we can say that the distance over the displacement is equaling the initial displacement, which we know to be zero plus the initial velocity times, time mocha. And this would be plus 1/2 a T squared. Now, in this case, we can say that here there isn't any get there isn't going to be any acceleration inthe e x direction, and we can say that then the time would be equaling the distance divided by the initial velocity of the truck. And so this is equaling 16 meters, divided by 9.50 meters per second, and this is equaling 1.68 seconds. So this is the time of flight. Just keep this in mind, and we found it by taking the displacement in the ex direction and dividing it by the initial velocity in the extraction. Given that we don't have any acceleration in the ex direction and so we can then say that here for party the angle that the boy throws the ball, we can say that with respect to the truck uh, the boy, the rose, the ball we could say, uh, straight up straight upwards so you can say zero degrees with the vertical or we can say 90 degrees from the horizontal. Either one is correct. 90 degrees from the horizontal would simply be 90 degrees. Uh, counterclockwise from the positive X axis or zero degrees with the vertical straight upwards allowed me your answer for party for part B. We want to find the initial speed of the can relative to the truck. So essentially, we could say that if that weakens, find the time, which would be we could say that dif the change in why position would be equaling V. Why initial times t plus 1/2 G. She squared and we know that the can is is going upwards and then coming back down. So it's it's ah, displacement in the wind direction is actually a zero because it's being launched from a certain level. And then it's coming back down to that same level, and we can then solve for the initial velocity in the UAE direction, which would simply be the initial velocity with respect to the truck. And so this would be equaling. 1/2 times 9.8 meters per second squared multiplied by 1.68 seconds. And this is equaling 8.23 meters per second. And so this would be our in the magnitude of our initial velocity, this would be upwards. And again, this would be with respect to the truck. And so we can say this would be your interview for a fee for part C. Um, what is it going to? What is the shape of the cans trajectory as seen by the boy? So, um, uh, as scene by the boy, uh, can will always be directly overhead, we can say going straight up, uh, then coming, uh, straight back down. So I would be your answer for part c and then for parte de um, we did this. We could say that it's a new observer on the ground. Watches the boy through the through the can now describe the shape of the trajectory. So for part D, we can say and ah, as seen from and observer on the ground, the cans trajectory, uh, will resemble a proble opening downwards. And what I mean by that is simply if this was the boy and let's just say this was the observer and the can goes up. This is exactly what the cans trajectory would look like. This is the movement of the truck right here. And then this would be the can going up and then coming straight down in the reference dream of the boy, it's always gonna be directly overhead. How about her For the Observer, you're going to see a trajectory. Are you gonna see a parabolic trajectory? And it's opening downwards. So that's what that's what we mean by a pair of a parabola opening downwards. And then for part E. We can find the essentially the initial velocity of the Cannes with respect to the Observer. And we can say that then the magnitude of the initial velocity would be equaling the square root of the ex velocity of the truck. 9.50 meters per second squared, plus the initial Y velocity of the can 8.23 meters per second quantity squared, and we find that the magnitude of the initial velocity from a stationary observer would be 12.5 meters per second and this would be in this quadrant one. So here they I only want the initial velocity. Um, so we can just say 12.5 meters per second. Um, however, it would be in the quadrant, runs quadrant one. So we have a positive why velocity and a positive X philosophy. That is what is being described in this trip in this ah parabolic trajectory for part deep. So this would be our final answer. And this is again with respect to A with respect to a stationary observer. That is the end of the solution. Thank you for watching.

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