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1)Suppose a metal sphere is launched up a ramp [incline of 20 degrees] with Vi = 1.5 m/s. The end of the ramp is 1.20 m
above the floor. Calculate the range of the sphere.
2)Now suppose that the ramp is tilted downwards as shown below. Suppose that the sphere leaves the ramp at 1.5 m/s. The bottom of the ramp is 0.90 m above the floor. Calculate the range of the sphere.
1) the ramp rises up and the ball moves up
2) the ramp goes down and the ball moves downwards.

2006-10-31 16:03:55 · 4 answers · asked by Neo 2 in Science & Mathematics Physics

4 answers

Now 2) makes a little more sense.

When the sphere leaves the bottom of the ramp, the horizontal component of its speed is 1.5 * cos 20.

The time taken for the sphere to fall to the floor is given by

.9 = (1/2) * 9.81 * t^2

around 3/4 of a second. Multiply the time by the horizontal component of the speed when it leaves the ramp and you will find how far from the bottom of the ramp the sphere hits the floor. Of course, the sphere will be spinning when it hits, and will roll away from the point of impact. How far it rolls depends on how much it is spinning, its moment of inertia, and the friction of the floor.

This is an idiotic question. Go back to the person who posed it and punch him in the nose.

2006-10-31 16:08:18 · answer #1 · answered by ? 6 · 0 0

You will need to find the moment of inertia of the sphere. This is a geometric factor based on its size and on whether it is solid or simply a hollow shell. It will make matters easier to visualize if you assume that the sphere is a particular size and of a particular mass. Given the moment of inertia, you can calculate the rotational kinetic energy in terms of the rolling speed, and using that plus the gravitational energy you can grind out the gory details.

2006-11-01 00:11:12 · answer #2 · answered by Anonymous · 0 0

Try Yahoo Education.

2006-11-01 00:06:24 · answer #3 · answered by KC 3 · 0 0

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