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. Show 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 ApproachWe 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:
\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 ApproachAssuming 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 MotionAs 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 EnergyNow, 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.
Updated March 13, 2018 By Petra Wakefield
Acceleration due to gravity causes a falling object to pick up speed as it travels. Because a falling object’s speed is constantly changing, you may not be able to measure it accurately. However, you can calculate the speed based on the height of the drop; the principle of conservation of energy, or the basic equations for height and velocity, provide the necessary relationship. To use conservation of energy, you must balance the potential energy of the object before it falls with its kinetic energy when it lands. To use the basic physics equations for height and velocity, solve the height equation for time, and then solve the velocity equation.
Ascertain the height from which the object fell. Multiply the height by the object's acceleration due to gravity. The acceleration due to gravity is 32.2 ft/s^2 for English units, or 9.8 m/s^2 for SI units. If you drop an object from 15 feet, for example, you would multiply 15 ft * 32.2 ft/s^2 to get 483 ft^2/s^2.
Multiply the result by 2. For example, 483 ft^2/s^2 * 2 = 966 ft^2/s^2.
Take the square root of the previous result to calculate the velocity when the object hits the ground. The square root of 966 ft^2/s^2 is 31.1 ft/s, so the object in this example would hit the ground traveling at 31.1 ft/s.
Ascertain the height from which the object fell. Multiply the height by 2, and divide the result by the object's acceleration due to gravity. If the object fell from 5 m, the equation would look like this: (2*5 m)/(9.8 m/s^2) =1.02 s^2.
Take the square root of the result to calculate the time it takes for the object to drop. For example, the square root of 1.02 s^2 equals 1.01 s.
Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.
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About the Author
Petra Wakefield is a writing professional whose work appears on various websites, focusing primarily on topics about science, fitness and outdoor activities. She holds a Master of Science in agricultural engineering from Texas A&M University. |
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