A steel ball rolls with constant velocity across a tabletop 0.950m high. IT rolls off and hits the ground +0.352m horizontally from the edge of the table. How fast was the ball rolling?

Question

A steel ball rolls with constant velocity across a tabletop 0.950m high. IT rolls off and hits the ground +0.352m horizontally from the edge of the table. How fast was the ball rolling?

Answer 1

y=vit+1/2at^2

-9.5=1/2(-9.81)t^2
t=0.44 s

x=vit
.352=vi(.44)
vi=0.8 m/s

Answer 2

To determine the speed at which the steel ball was rolling, we need to apply the principles of projectile motion. We can use the kinematic equations to calculate the initial horizontal velocity of the ball.

Let's break down the information given:
- The height of the tabletop is 0.950m.
- The ball lands on the ground 0.352m horizontally from the edge of the table.

Considering the horizontal motion, we can assume that the acceleration in this direction is zero. Therefore, the initial horizontal velocity of the ball, Vx, will also be its final horizontal velocity just before it rolls off the table.

We can use the equation for horizontal displacement:

Horizontal displacement (Δx) = Vx * time

Since the ball rolls with constant velocity across the tabletop, the time it takes to reach the edge is the same as the time it takes to reach the ground. We can calculate the time using the vertical motion equation:

Vertical displacement (Δy) = Vy * time + (1/2) * (-g) * time^2

The initial vertical velocity of the ball, Vy, is zero because the ball is rolling horizontally. The term (-g) represents the acceleration due to gravity, which is approximately -9.8 m/s^2. The vertical displacement, Δy, is the height of the tabletop, which is 0.950m.

Simplifying the vertical motion equation, we get:

0.950m = (1/2) * (-9.8 m/s^2) * time^2

Solving for time, we find:

time = √(2 * 0.950m / 9.8 m/s^2)

Substituting this time value back into the horizontal displacement equation, we have:

0.352m = Vx * (√(2 * 0.950m / 9.8 m/s^2))

Now we can solve for Vx, the horizontal velocity:

Vx = 0.352m / (√(2 * 0.950m / 9.8 m/s^2))

Calculating this expression will give us the velocity at which the ball was rolling before falling off the table.

Answer 3

Why did the steel ball leave the table? It just couldn't resist the pull of gravity, it wanted to be a baller on the ground! Now, to answer your question, we can use some good old physics. Since the velocity is constant, we can assume there were no forces acting on the ball horizontally. So, using the horizontal distance of 0.352m, we can calculate the time it took for the ball to reach the ground. Then, we can use this time to find the initial velocity of the ball. But remember, this calculation is no joke!

Answer 4

Get the time to fall from the height, H = 0.950 m. Call that time T. You will need to use the acceleration of gravity (g) to compute T.

H = (g/2) T^2
Haven't you seen that equation before for falling bodies?

Then use the relationship
V T = 0.352
to get V

Answer 5