2.5.2 Height and Range

When people launch projectiles, they often obsess over how high H they reach and how far R they land.

Height

Remember that a projectile motion is like a vertical throw that happens to move side way too. So the maximum height of a projectile H has nothing to do with the horizontal motion. In fact, it is dependent only on its initial vertical component of velocity u_y, (and g, if you insist) and can be calculated in just one equation.

$\begin{aligned}({{v}^{2}}&={{u}^{2}}+2as)\\0&={{u}_{y}}^{2}-2gH\\H&=\frac{{{{u}_{y}}^{2}}}{{2g}}\end{aligned}$

Time of flight

How long the projectile stay in the air before landing has nothing to do with the horizontal motion. In fact, the time-of-flight t_f is dependent only on the initial vertical component of velocity u_y. Using the fact that the vertical velocity merely switches direction upon landing, we can do

$\begin{aligned}(v&=u+at)\\-{{u}_{y}}&={{u}_{y}}-g{{t}_{p}}\\{{t}_{p}}&=\frac{{2{{u}_{y}}}}{g}\end{aligned}$

Alternatively, you can mentally figure out that the time to reach the peak is ${{t}_{p}}=\frac{{{{u}_{y}}}}{g}$ . And since it takes as long to go up as come down, $\displaystyle {{t}_{f}}=2{{t}_{p}}=\frac{{2{{u}_{y}}}}{g}$

Range

The horizontal range,

$\displaystyle \begin{aligned}R&=\text{horizontal speed }\!\!\times\!\!\text{ time of flight}\\&={{u}_{x}}\times {{t}_{f}}\\&={{u}_{x}}\times \frac{{2{{u}_{y}}}}{g}\end{aligned}$

Notice that the horizontal range is dependent on both the vertical and horizontal initial velocities u_x and u_y. While u_y decides how much time is available to the projectile to move forward, u_x decides how fast it is moving forward during that time.

Too high a launch angle θ is no good because u_x will be too small. Too low a launch angle θ is also no good because u_y (and thus t_f) will be too small. So, what is the optimal angle?

Let’s expand the terms and let the math show us.

$\begin{aligned}R&=\frac{{2{{u}_{x}}{{u}_{y}}}}{g}\\&=\frac{{2(u\cos \theta )(u\sin \theta )}}{g}\\&=\frac{{{{u}^{2}}\sin 2\theta }}{g}\end{aligned}$

Since $\sin 2\theta$ ranges from 0 to 1 (for $0{}^\circ \le \theta \le 90{}^\circ$ ), the maximum horizontal range is

$\displaystyle {{R}_{{\max }}}=\frac{{{{u}^{2}}}}{g}$

This occurs when $\sin 2\theta =1$ , which occurs when $\theta =45{}^\circ$ . So for a projectile with a fixed launch speed u, 45° is the optimal angle to launch it at, if maximum range is desired.

Example

A cannon ball is fired at 120 m s^-1 at an inclination angle of 25°. Ignoring the effect of air resistance, calculate

a) the speed of the cannon ball 8.0 seconds later,

b) the distance between the cannon ball from the launch position 8.0 seconds later,

c) the maximum height reached,

d) the landing position of the cannon.

Solution

$\begin{aligned}{{u}_{x}}&=120\cos 25{}^\circ &=108.8\text{ m }{{\text{s}}^{{-1}}}\\{{u}_{y}}&=120\sin 25{}^\circ &=50.71\text{ m }{{\text{s}}^{{-1}}}\end{aligned}$

$\displaystyle \begin{aligned}({{v}_{y}}&={{u}_{y}}+{{a}_{y}}t)\\{{v}_{y}}&=50.71-(9.81)(8.0)&=-27.77\text{ m }{{\text{s}}^{{\text{-1}}}}\end{aligned}$

$\begin{aligned}\left| v \right|&=\sqrt{{{{{108.8}}^{2}}+{{{27.77}}^{2}}}}\\&=112\,\text{m }{{\text{s}}^{{-1}}}\end{aligned}$

$\begin{aligned}({{s}_{x}}&={{u}_{x}}t)\\{{s}_{x}}&=(108.8)(8.0)&=870.4\text{ m}\end{aligned}$

$\begin{aligned}({{s}_{y}}&={{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}})\\{{s}_{y}}&=(50.71)(8.0)+\frac{1}{2}(-9.81){{(8.0)}^{2}}&=91.76\text{ m}\end{aligned}$