i have homework problem in physics and i am having hard time to get the solution...so anyone who can help?

Question

a chemistry student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 11 m/s. The student throws a ball into the air on a trajectory that she observes to make an initial angle of 54degree with respect to the horizontal along the same line as the track. The a physics student, who is standing on an embankment nearby, observes the ball to rise straight up vertically.

How high does the ball rise? (What is the horizontal component of the velocity of the ball as measured by the student?)

Answer 1

Note that , in order for the ball's trajectory to be seen to be
vertical by a stationary observer, the ball's horizontal component of velocity must be equal to and opposite in direction of that of the moving train.
So the ball is thrown -11m/s w.r.t. the train. At 54 degrees from the horizontal, the vertical component is V where tan(54)=V/11.
Hence V=11tan(54)=(11)1.376=
15.14m/s.
So with V^2=2aS, (velocity, acceleration,distance respectively)
S=V^2/2a=
(15^2m^2/s^2)/((2)9.75m/s^2))=
11.75 m high

Answer 2

Couple of thoughts: 1. The squares and the polygons have equal areas, or 2. The squares and the polygons have equal anglular areas based on the center of the polyhedron, or 3. The centers of the squares and the polygons are equal distance from the center of the polyheadron, or 4. The squares and the polygons are set a distance from the center such that, when they are are rest, the will have equal potential energies. I like the last one best. This means there is no 'hole' to make it more likely that a square or a polygon would trap the die. I am also concerned (and get stuck at) modeling how this thing rolls. Things like spin and surface friction are important. I suspect, however, that it would end up rolling with the edges aligned. I.e., pick an edge of a square. Its final rotation would be in a direction perpendicular to that edge. This type of rotation would require the least kinetic energy to maintain. I note that this type of rotation would pass over 2 squares and 4 polygons so the squares might need to be slightly bigger to catch an appropriate share of the die. ..................... I'd bet that the answer would lie some place close to the four thoughts mentioned above [which themselves should be fairly close together]. I think it would come down to a trial and error engineering problem. >>>>>>>>>>> I have thought about this some more. I am pretty sure the die would end up rolling the way I described. Square- polygon -polygon- Square- polygon- polygon. I then note that the angles for the Square polygon is 62 degress and the plygon-polygon is 54 degrees. I'd further assume that the moment of interia is as if a ball and a constant resitive force. I would then look at the energy necessary to get over each angle and size everything so a square lands 3 times for every 4 polygons. *Note, it won't land on the polygon's evenly, there will be a tendency to land on the second polygon Then it is off to make a grant proposal for funding because there are simply too many variable and unknowns. >>>>>>>>>> Just buy an 8-sided die and blacken one side. That should work as a 7-side die. >>>>>>>>>> Duke's method inspired me. Just take a sphere and grind down 14 points equally (to form circles), with each circle centered where the square or the polygon would be. Leave a little room between the circles. It would be close to being perfectly fair. It would also look good.

Answer 3

The student on the embankment sees it as having 0 horizontal velocity since it goes straight up. Therefore to the other student, the horizontal velocity is 11 m/s in the direction opposite to the train's travel. Using a tan54 you can get the vertical velocity, Vv. And then you can get the height, y, from
V^2 = Vv^2 +2gy if V = 0, which it does at the top of its trajectory.

Answer 4

1

Answer 5

HEre are some hints, If the ball appears to rise straight up to the observer, the only way this is possible is if the relative velocity in the horizontal direction is zero. This menas she must throw the ball in such a way that its horizontal velocity is directly equal and opposite of the velocity of the car that she is on. knowing this information and the angle along with some trigonometry should give you the vertical velocity, from which you can calculate the maximum hieght the ball travels.

Answer 6

Part 2: For the ball to have 0 horizontal velocity with respect to the physics student (it rises straight up), it must have -11 m/s velocity with respect to the girl.

Part 1:
Let v = vertical velocity of the ball.
v = 11*tan54
(we no longer care about horizontal direction)
v^2-0 =2as (multiplied equation by -1 for readability)
(s = height we're looking for)
s = ((11tan54)^2)/(2a)

Make sure you have dimensional consistency before cranking out the number.

Answer 7

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Answer 8

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Answer 9

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