# Velocity of a Ball When it Hits the Ground

If a ball is thrown vertically upward from the roof of a 32 ft. tall building with a velocity of 80 ft/sec, it's height in feet after t seconds is s(t)=32+80t-16t^2

I know the maximum height is 132ft

What is the velocity of the ball when it hit's the ground (Height 0)?

• Well, the first question you should ask yourself is how velocity relates to position (as a function of time). – Seth May 4 '14 at 18:47

Due to symmetry (you can verify this with calculation), the ball will be moving 80 ft/sec when it comes back down to the point from which you threw it on the building. So you can reduce the problem to describing a ball thrown off a building with initial velocity 80 ft/sec downward. We have

$$s(t) = 32 - 80 t - 16 t^2,$$

as the ball's position function of time. To find $t$ at $s=0$, we may use the quadratic formula, which gives

$$t = \frac{80 \pm \sqrt{80^2+4(16)(32)}}{-32},$$

which simplifies to approximately $t=0.37$. Since acceleration is constant, we have

$$v(t) = -80 - 32t,$$

as the ball's velocity function of time. Solving for $v$ when $t=0.37$, we obtain $v=-91.84$, which seems reasonable enough. The negative sign comes from calling the upwards direction $+$ and the downwards direction $-$.

• Thank you, can you explain how you got v(t)=-80-32t? Is it the derivative of the original equation? – Natalie May 4 '14 at 19:01
• The formula that I used was $v(t)=v_0 + a\cdot t$, which is true when $a$ is a constant. Since $s(t)$ tells us that $\frac{a}{2}=-16$, we know that $a=-32$. Then, we can use the fact that the ball's velocity at the top of the building is $-80$ to give a $v_0$. Plug in $t$ from before and you get the answer. – alethiometryst May 4 '14 at 19:06
• You could get it by differentiating $s(t)$, which of course would be the faster way to go about it. :) – alethiometryst May 4 '14 at 19:07
• Got it, thank you everyone <3 – Natalie May 5 '14 at 0:54

If you know the maximum height, the answer is really simple to find - we can directly work from there, knowing that the velocity at maximum height is $0 \frac m s$ because it is, essentially, not moving as it turns from upwards motion to downwards free-fall.

Note that there are two approaches to finding this solution - one, more basic and perhaps easier to understand (but longer and more complex), the kinematic approach, using basic ideas of velocity, acceleration, and displacement - and the other, although more advanced and, much simpler - the energy approach. I'll start with the basic solution using the kinematic approach.

## 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.

• It may be worth while to look into Stack Exchange's Physics site. – Ephraim May 8 '14 at 4:17
• Point; I'll jump into that and have my fun. – Kye W Shi May 8 '14 at 4:33

Since the (vertical) velocity of the ball is given by (assuming $g \approx$ constant)

$$v(t) = \frac{ds}{dt} = v_0 - g t,$$

the speed of the object when hits the ground is the velocity at time $t^*$, where $t^*$ is the value of $t$ when $s(t) = s_G = 0$ (if your reference system is placed at $s = s_G = 0$ for the ground). Find $t^*$ and substitute back in the expression of $v$.

Cheers.

• I am confused on how to find the time? Is it just the derivative of 32+80t-16t^2? – Natalie May 4 '14 at 18:54

The position of the ball $x$ is a function of time: $$x(t)=32+80t-16t^2$$ Let $t^*$ be the time at which the ball hits the ground. In math lingo, this means $x(t^*)=0$, and so we can solve this algebraic equation for $t^*$ (hint: know how to solve a quadratic?). The velocity of the ball is also a function of time: $$v(t)=D_tx(t)=80-32t$$ So the velocity of the ball when it hits the ground is, by definition, $v(t^*)$, assuming you know $t^*$, which you do (read the bolded message).

Since the velocity of the ball is $0 \frac{ft}{s}$ when the ball is at its maximum height of $132ft$, you can calculate it starting from there. Since this is more of a physics problem than a math problem, I'm going to do this the more Physics-y way. To make it easier, I'm converting this to meters. $132ft = 40.2336m \approx 40m$. Starting from the momentum principle: $$\Delta P = F_{net}*\Delta t$$ $$P_{final} = P_{initial} + F_{net}\Delta t$$ Since we are starting when the velocity of the ball is $0 \frac{m}{s}$, $P_{initial} = 0$, so : $$P_{final}=\Delta P = F_{net}\Delta t$$ $$P = mass* velocity = m*v$$ $$m*v_{final} = F_{net}\Delta t$$ $$F_{net} =mass*acceleration = m*a$$ $$m*v_{final} = (m*a)\Delta t$$ $$v_{final} = \frac{m*a*\Delta t}{m} = a*\Delta t$$ $$v = \frac{distance}{time} = \frac{r}{\Delta t}$$ $$\frac{r}{\Delta t} = a*\Delta t$$ $$r = a*\Delta t^2 = 40 meters$$ $$\sqrt{\frac{r}{a}} = \Delta t$$ since $v$ is actually the derivative of $r$ (because velocity is the rate of change of position) $$\frac{d}{dt}r = \frac{d}{dt}a\Delta t^2 = 2a\Delta t = v$$ $$\therefore v = 2a\Delta t = 2a\sqrt{\frac{r}{a}} = \frac{2\sqrt{a}^2\sqrt{r}}{\sqrt{a}} = 2\sqrt{a}\sqrt{r} = 2\sqrt{a*r}$$ on earth, $a=g=9.8$ so $$v=2\sqrt{9.8*40} = 2\sqrt{392} = \boxed{28\sqrt{2}\frac{m}{s}\approx 129.9\frac{ft}{s}}$$