Why Does the Ball Appear to Rise Vertically to the TA?

[Juntao]

A Physics 111 student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 12 m/s. The student throws a ball into the air on a trajectory that she observes to make an initial angle of 53° with respect to the horizontal along the same line as the track. The student's TA, who is standing on an embankment nearby, observes the ball to rise straight up vertically.

How high does the ball rise in meters?

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Of course, I tried this equation, (vy)^2=(v0 sin theta)^2-2(g)(y-y0)
where vy is final velocity in y component
v0=initial velocity
g=9.8 m/sec/sec
y0=0

So if the ball reaches top of its height, its velocity is zero, thus
0=(12*sin 53)^2-(2*9.8)(y)
and I get y=4.69 meters.

Okay, this isn't right, and I'm probably going at this problem the wrong way, right? Argh!

 

[HallsofIvy]

It looks to me like you are assuming vo= 12 m/s and there is no reason to do that. What you are told is that the ball rises vertically relative to the ground. Since the train is moving at 12 m/s, the HORIZONTAL COMPONENT of the ball (relative to the train) must be 12 m/s. Use that to find vo.
(The student is, by the way, throwing the ball opposite to the direction of the train.)

 

[M98Ranger]

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It seems like you are on the right track, but there are a few things you may want to consider. First, the equation you used is for the vertical motion of the ball, but the train is also moving horizontally. This means that the ball will have a horizontal velocity as well, and its trajectory will be a combination of vertical and horizontal motion. You may want to consider using the equations for projectile motion, which take into account both horizontal and vertical components of motion.

Additionally, it's important to keep in mind that the ball is thrown from a moving train, so its initial velocity will be a combination of the train's velocity and the velocity at which the student threw the ball. This may affect the angle at which the ball is thrown and its initial velocity.

Lastly, it's important to check your units. In your equation, you used a value of 9.8 m/sec/sec for acceleration due to gravity, but the units for this value should be m/s^2. Double checking your units can help ensure that your final answer is in the correct units as well.

I hope this helps and good luck with your problem!