Which of the following is the largest when the height attained by the projectile is the greatest

The maximum height calculator is a tool for finding the maximum vertical position of a launched object in projectile motion. Whether you need the max height formula for an object starting directly off the ground or from some initial elevation - we've got you covered. If you're still wondering how to find maximum height of a projectile, read the two short paragraphs below, and everything should become clear.

Maximum height of the object is the highest vertical position along its trajectory. The object is flying upwards before reaching the highest point - and it's falling after that point. It means that at the highest point of projectile motion, the vertical velocity is equal to $0$ (vy=0v_y = 0vy​=0).

0=vy–g⋅t=v0⋅sin⁡(α)–g⋅th0 = v_y – g \cdot t = v_0 \cdot \sin(\alpha) – g \cdot t_{\mathrm{h}}0=vy​–g⋅t=v0​⋅sin(α)–g⋅th​

From that equation we can find the time tht_{\mathrm{h}}th​ needed to reach the maximum height hmaxh_{\mathrm{max}}hmax​:

th=v0⋅sin⁡(α)gt_{\mathrm{h}} = v_0\cdot\frac{\sin(\alpha)}{g}th​=v0​⋅gsin(α)​

The formula describing vertical distance is:

y=vy⋅t−g⋅t22y = v_y\cdot t - g\cdot \frac{t^2}{2}y=vy​⋅t−g⋅2t2​

So, given y=hmaxy = h_{\mathrm{max}}y=hmax​ and t=tht = t_{\mathrm{h}}t=th​, we can join those two equations together:

hmax=v0⋅th−g⋅th22=v02⋅sin⁡2(α)g−g⋅(v0⋅sin⁡(α)g)22=v02⋅sin⁡2(α)2⋅g\begin{split} &h_\mathrm{max} = v_0\cdot t_\mathrm{h} - g\cdot\frac{t_\mathrm{h}^2}{2}\\ &=v_0^2\cdot \frac{\sin^2(\alpha)}{g} - g\cdot\frac{\left(v_0\cdot\frac{\sin(\alpha)}{g}\right)^2}{2}\\ &=v_0^2\cdot \frac{\sin^2(\alpha)}{2\cdot g} \end{split}​hmax​=v0​⋅th​−g⋅2th2​​=v02​⋅gsin2(α)​−g⋅2(v0​⋅gsin(α)​)2​=v02​⋅2⋅gsin2(α)​​

And what if we launch a projectile from some initial height $h$`? No worries! Apparently, the calculations are a piece of cake - all you need to do is add this initial elevation!

hmax=h+v02⋅sin⁡(α)2⋅gh_\mathrm{max}= h+\frac{v_0^2\cdot \sin(\alpha)}{2\cdot g}hmax​=h+2⋅gv02​⋅sin(α)​

Let's discuss some special cases with changing angle of launch:

• If $\alpha =90°$, then the formula simplifies to:

hmax=h+v022⋅gh_{\mathrm{max}} = h+\frac{v_0^2}{2\cdot g}hmax​=h+2⋅gv02​​

And the time of flight is the longest.

If, additionally, vy=0v_y = 0vy​=0, then it's the case of free fall, which we detailed at the free fall calculator. Also, you may want to have a look at our even more accurate equivalent - the free fall with air resistance calculator.

• If $\alpha =45°$, then the equation may be written as:

hmax=h+v024⋅gh_{\mathrm{max}} = h+\frac{v_0^2}{4\cdot g}hmax​=h+4⋅gv02​​

And in that case, the range is maximal if launching from the ground ($h=0$).

• If $\alpha =0°$, then vertical velocity is equal to $0$ (vy=0v_y = 0vy​=0), in this case we can calculate the horizontal projectile motion. As sine of $0°$ is $0$, then the second part of the equation disappears, and we obtain :

hmax=hh_\mathrm{max} = hhmax​=h

The initial height from which we're launching the object is the maximum height in projectile motion.

The motion of a projectile is a classic problem in physics, and it has been analyzed under every possible aspect. The fact that we can easily reproduce it and observe it was a contributing factor. We decided to create a suit of tools related to the motion of a projectile:

Just relax and look how easy-to-use this maximum height calculator is:

1. Choose the velocity of the projectile. Let's type $30ft/s$.
2. Enter the angle. Assume we're kicking a ball ⚽ at an angle of $70°$.
3. Optionally, type the initial height. In our case, our starting position is the ground, so type in $0$. Can the ball fly over a $13\mathrm{ft}$ fence?
4. Here it is - maximum height calculator displayed the answer! It's $12.35\mathrm{ft}$. So it will not fly over the mentioned barrier - throw it harder or increase the angle to reach your goal.

Just remember that we don't take air resistance into account!