Question

A ball thrown vertically with a velocity of 30 m/s. What is the velocity at the top of its path?

What is its maximum height?

Answer 1

The maximum height is `46"m"`, and the velocity at this point is momentarily `0"m"//"s"`.

Explanation:

We are given the following information:

• `v_i=30" m"//"s"`

Which doesn't seem like much! But we can actually get three more pieces of information by thinking about the motion of the ball.

For instance, we know that (ignoring air resistance, etc) the ball is under only the influence of gravity when it leaves the person's hand. Therefore, the acceleration of the ball is equal to the free-fall or gravitational acceleration constant: `-g`.

Next, we know that at the ball's maximum altitude it will momentarily stop as it changes direction before falling back down. This means that at the very top of the motion, `v=0`.

Additionally, we can define the starting position at `y=0`.

Now we have:

• `v_i=30" m"//"s"`
• `v_f=0" m"//"s"`
• `y_i=0"m"`
• `a=-9.81"m"//"s"^2`

We have three kinematic equations to choose from:

• `v_f=v_i+aDeltat`
• `v_f^2=v_i^2+2aDeltas`
• `s_f=s_i+v_iDeltat+1/2a(Deltat)^2`

To find the maximum height of the ball, we can use the only kinematic equation available to us which does not involve time, as we do not have that variable.

`color(blue)(v_f^2=v_i^2+2aDeltas)`

As we know that the final velocity is zero:

`=>cancel(v_f^2)=v_i^2+2aDeltas`

`=>0=v_i^2+2aDeltas`

We can solve for `Deltas`:

`color(blue)(Deltas=(-v_i^2)/(2a))`

Using our known values:

`Deltas=(-(30"m"//"s")^2)/(2*-9.81"m"//"s"^2)`

`=>=45.87"m"`

`color(blue)(Deltas=46"m")`