Question

A baseball is rolling along a tabletop with a velocity of 3.9 m/s to the right. The tabletop is 1.1 m above the ground. The ball rolls off the edge of the table and falls to the ground.How far from the table does the ball land?

Answer 1

Let's name some variables that we know and need:

y: height; y = 1.1 m

g: gravitational acceleration; g = 9.81 m/s^2

t: time in seconds (we need to solve for this)

Given these variables, we can use the following equation to solve for t.

`y=(gt^2)/(2)`

Now, plugging in the variables we know,

`1.1=(9.81t^2)/(2)`

And solving for t,

t = 0.4736 s

This is how long it takes for the ball to reach the ground after it falls off the edge.

Since there is no horizontal acceleration, the ball continues travelling horizontally at 3.9 m/s.

Let's name some more variables:

x: horizontal distance travelled after falling off the table (we need this)

v: horizontal velocity; v = 3.9 m/s

We can use this equation to solve for x:

`x=vt`

Plugging in for v and t,

`x=3.9*0.4736`

Finally,

x = 1.84704 m