# A projectile is shot at a velocity of 2 m/s and an angle of pi/12 . What is the projectile's maximum height?

Jul 13, 2017

$\text{maximum height} = 0.0137$ $\text{m}$

#### Explanation:

We're asked to find the maximum height reached by a projectile given its initial speed and launch angle.

At its maximum height, the object's instantaneous $y$-velocity is $0$. We can use the equation

${\left({v}_{y}\right)}^{2} = {\left({v}_{0 y}\right)}^{2} + 2 {a}_{y} \Delta y$

where

• ${v}_{y}$ is the $y$-velocity at a certain height $\Delta y$. This value is $0$ because we're trying to find its maximum height.

• ${v}_{0 y}$ is the initial $y$-velocity.

We find this using the equation

${v}_{0 y} = {v}_{0} \sin {\alpha}_{0}$

where

• ${v}_{0}$ is the initial speed (given as $2$ $\text{m/s}$)

• ${\alpha}_{0}$ is the launch angle, given as $\frac{\pi}{12}$

Therefore,

v_(0y) = (2color(white)(l)"m/s")(sin(pi/12)) = color(red)(0.518 color(red)("m/s"

• ${a}_{y}$ for any object in free-fall is $- g$, equal to $- 9.81$ ${\text{m/s}}^{2}$

• $\Delta y$ is the height (relative to original height, which is taken to be $0$), which is what we're trying to find

A list of variables we're using:

${v}_{y} = 0$

v_(0y) = color(red)(0.518 color(red)("m/s"

${a}_{y} = - 9.81$ ${\text{m/s}}^{2}$

Deltay = ?

Plugging in known values into the equation, we have

0 = (color(red)(0.518color(white)(l)"m/s"))^2 - 2(9.81color(white)(l)"m/s"^2)(Deltay)

$\left(19.62 \textcolor{w h i t e}{l} {\text{m/s}}^{2}\right) \left(\Delta y\right) = 0.268$ ${\text{m"^2"/s}}^{2}$

Deltay = (0.268color(white)(l)"m"^2"/s"^2)/(19.62color(white)(l)"m/s"^2) = color(blue)(0.0137 color(blue)("m"

The maximum height reached by this projectile is thus color(blue)(0.0137 sfcolor(blue)("meters", which makes some sense because it was launched (a) with a small initial speed and (b) with a low angle relative to the horizontal (equivalent to ${15}^{\text{o}}$).