How Fast Was the Ball Going When It Left the Incline?

[kofmelk]

Homework Statement

A ball rolls off an incline plane on top of a 1m high table and strikes the floor 62.5cm from the edge of the table. What was the velocity of the ball just as it left the incline plane?

The incline is given as 30 degrees.

I've put my frame of reference at the point where the ball is at the edge of the table about to fall off.

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Homework Equations

1. Position final = Position initial + (Velocity initial * time) + (1/2 Acceleration * time^2)
2. Velocity final = Velocity initial + (Acceleration * time)

The Attempt at a Solution

Since my frame of reference is placed where it is, in the x direction the acceleration is 0 as well as the initial position. So formula 1 in the x-direction boils down to:

.625m = (Velocity initial * time)

Or in other words:

Vi = .625/t

I tried to solve for t in the y-direction but that's where I got stuck. It travels down the ramp first so its initial velocity and position as well as the final velocity are unknown.

Stuck. Thanks for any and all help.

 

[tiny-tim]

call the initial speed v, and split it into x and y components

its initial velocity and position as well as the final velocity are unknown

No.

Its initial position is x = 0, y = 1.

Its initial speed is unknown, so call it v.

Its initial direction is known: 30º to the x-axis.

Split the velocity into x and y components.

Do the y-direction first, and work out the time to reach y = 0 (it'll involve v).

Then use that time to work out the x distance traveled in that time. I'll involve v, and you can put it equal to 0.625!

 

[Moetasim]

I would approach this problem by first identifying the key variables and equations that can be used to solve for the unknown velocity of the ball. The key variables in this problem are the height of the table (1m), the distance the ball travels before hitting the ground (0.625m), and the angle of the incline (30 degrees). The equations that can be used are those of kinematics, which describe the motion of objects.

To solve for the velocity of the ball, we can use the equation for displacement in the x-direction:

x = x0 + v0x*t + (1/2)*ax*t^2

Where x0 is the initial position, v0x is the initial velocity in the x-direction, ax is the acceleration in the x-direction (which is 0 in this case), and t is the time. Since we have set our frame of reference at the edge of the table, x0 = 0 and ax = 0. Therefore, the equation becomes:

x = v0x*t

Substituting in the given values, we get:

0.625m = v0x*t

Next, we can use the equation for displacement in the y-direction:

y = y0 + v0y*t + (1/2)*ay*t^2

Where y0 is the initial position, v0y is the initial velocity in the y-direction, and ay is the acceleration in the y-direction (which is due to gravity and is equal to -9.8 m/s^2). Again, since our frame of reference is at the edge of the table, y0 = 0. The final position, y, is equal to the height of the table (1m). Therefore, the equation becomes:

1m = v0y*t + (1/2)*(-9.8 m/s^2)*t^2

Simplifying, we get:

1m = v0y*t - 4.9*t^2

Substituting in the known value for t from our previous equation, we get:

1m = v0y*(0.625m/v0x) - 4.9*(0.625m/v0x)^2

Solving for v0y, we get:

v0y = 1m/(0.625m/v0x) - 4.9*(0.625m/v0x