[Math] how to find the average velocity and instantaneous velocity

calculus

This is a question from my Calculus book. I was showed the correct answer but I really need to learn what the process is to achieve the answer. Here is the question:

If a ball is thrown into the air with a velocity of $40 \frac{ft}{s}$, its height in feet $t$ seconds later is given by $y=40t – 16t^2$ . Find the average velocity for the time period beginning when $t = 2$ and lasting

(i) 0.5sec

(ii) 0.1sec

(iii) …

Best Answer

First you find the height after 2 seconds, we'll denote that as $y_2$. So using the fomula we have:

$$y_2 = 40 \cdot 2 - 16 \cdot 2^2 = 80 - 64 = 16$$

After 0.5 seconds, the time passed after the throw will be 2.5 seconds so for this height we have:

$y_{2.5} = 40\frac 52 - 16 \left(\frac 52\right)^2 =100 - 100 = 0$$

So after 2.5 second the height of the ball will be $0\text{ ft}$.

We know that the average velocity can be calulated using the following formula:

$$v = \frac{\Delta s}{\Delta t} = \frac{|y_1 - y_0|}{|t_1 - t_0|}$$

So after the substitution we have:

$$v = \frac{|0 - 16|}{|2.5 - 2|} = \frac{|-16|}{|0.5|} = \frac{16}{0.5} = 32$$

So the average speed in that period will be 32 $\frac{ft}{s}$

Now you can do the 0.1 second interval by yourself.

I. The Kinematic Approach

We know that, for any object under constant acceleration, the velocity at any point in time can be modeled by

$$v = v_0 + at$$

As previously stated, because the ball falls starting from its maximum height, the initial velocity is $0$. Therefore, the equation becomes

$$v = at$$

where $v_0$ is the initial velocity, $a$ is acceleration, and $t$ is the time elapsed. For most free fall cases, $a = -g$ (where $g$ is gravitational acceleration) as gravitational acceleration is downwards. However, in this particular case, we already know the maximum height and only need to examine the part of the trajectory that is downwards - therefore, if we simply "rotate our axes," setting the direction of motion as the positive direction, we can set $a = g$ because gravity would be working in the same direction of our current motion. Therefore, we can modify the equation to become

$$v = gt$$

Now all that's left is finding the time elapsed to reach the ground. We know, first of all, that we can find the displacement of an object under constant acceleration by averaging the initial and final velocities and multiplying that by time, thus

$$\Delta x = \frac {t(v_0 + v)} 2$$

Again, here, we know $v_0 = 0$, so the equation becomes

$$\Delta x = \frac {t(v)} 2$$

From here, we can solve for $t$ and then substitute our solution for $t$ into the previous equation to solve for $v$.

\begin{align*} \Delta x &= \frac {t(v)} 2 \quad && \\ \frac {2\Delta x} v &= t \quad && \text{Multiply both sides by $\frac 2 v$} \\ t &= \frac {2\Delta x} v \quad && \text{Solve for $t$} \end{align*}

Then, we substitute that into the previous equation:

\begin{align*} v &= gt \quad && \text{Previous equation} \\ v &= g(\frac {2\Delta x} v) \quad && \text{Substitute for $t$} \\ v &= \frac {2g\Delta x} v \quad && \text{Simplify fractions} \\ v^2 &= 2g\Delta x \quad && \text{Multiply both sides by $v$} \\ v &= \sqrt{2g\Delta x} \quad && \text{Square root both sides} \end{align*}

Then, we can plug in everything we already know to solve for $v$. Because of personal preferences ;), I'll do it in SI units:

$g = 9.81 \frac m {s^2}$ or $32.19 \frac {ft} {s^2}$, the gravitational acceleration
$\Delta x = 132 ft$ or $40.23 m$, the maximum height (or total displacement)

\begin{align*} v &= \sqrt{2 \cdot 9.81 \frac m {s^2} \cdot 40.23 m} \\ &\approx 28.04 \frac m s \\ &\approx 91.99 \frac {ft} s \text{ or } 92 \frac {ft} s \end{align*}

II. The Energy Approach

Assuming a system consisting of just the ground (gravitational force) and the ball, we can apply the Law of Conservation of Energy, which states that, in any closed system, the net energy (kinetic + potential) is constant. Here, we need only to find the energy at the two endpoints of motion and equate the two.

Energy at Peak of Motion

As previously stated, the velocity is 0 at the peak of motion - therefore, the only energy present would be potential energy. We know that gravitational potential energy is given by $mgh$, where $m$ is the mass of the ball, $g$ is the gravitational acceleration, and $h$ is the height. Thus

$$E = mgh$$

Energy at Ground (Bottom)

Here, when the ball is at the ground, the height $h$ is essentially $0$ (except that tiny (negligible) bit, as it isn't quite in contact with the ground yet), and all that needs to be measured is kinetic energy. We know that kinetic energy is given by $\frac 1 2 mv^2$, therefore

$$E = \frac 1 2 mv^2$$

Conservation of Energy

Now, we can easily equate the energy representations at different points in the ball's trajectory at the two endpoints and solve for $v$.

\begin{align*} mgh &= \frac 1 2 mv^2 \quad && \text{Equate the two energy representations} \\ gh &= \frac 1 2 v^2 \quad && \text{Divide both sides by $m$} \\ 2gh &= v^2 \quad && \text{Multiply both sides by 2} \\ \sqrt{2gh} &= v \quad && \text{Square root both sides} \\ v &= \sqrt{2gh} \quad && \text{Solve for $v$} \end{align*}

Unsurprisingly, this yields the exact same result as it would with the kinematics approach - but is a much shorter and more concise solution - the only downside to this solution is that it requires a deeper understanding of Physics concepts of work and energy.

[Math] Height and velocity of ball thrown vertically

You made a minor mistake: $v(t)$ should be $112 - 32t$. This should make everything correct. As for the second part, ground means $s(t)=0$. So find the corresponding time and plug into $v(t)$ to find the impact velocity.