# A projectile is shot from the ground at a velocity of 13 m/s and at an angle of (5pi)/12. How long will it take for the projectile to land?

Jan 13, 2016

The time a projectile is in the air is determined by the vertical displacement, initial vertical velocity, and the acceleration.

#### Explanation:

While a projectile problem is usually a two dimensional one, the air time is dependent only on the vertical component, so we can ignore the horizontal.

Our possible variables for an acceleration problem are:
${\vec{v}}_{i}$
${\vec{v}}_{f}$
$\vec{a}$
$\Delta \vec{d}$
$\Delta t$
And we need any three to find a fourth. W are looking for $\Delta t$, so we need any three of the others.

Though it is not stated in the question, one can assume from context that the projectile starts and ends on the ground. If this is the case, then start and end height are the same, so $\Delta \vec{d} = 0$.
Vertical acceleration is provided by gravity, so $\vec{a} = - 9.8 \frac{m}{s} ^ 2$.

With two variables down, we just need one more. We can use information about the launch velocity vector to find the vertical component of the initial velocity.

Given the parameters, we can get the vertical component of the initial velocity using

${v}_{i y} = {v}_{i} \sin \left(\theta\right)$

where ${v}_{i} = 13 \frac{m}{s}$ and $\theta = \frac{5 \pi}{12}$, so

${v}_{i y} = 13 \sin \left(\frac{5 \pi}{12}\right) = 12.56 \frac{m}{s}$

Armed with $\Delta \vec{d} , \vec{a} ,$ and ${\vec{v}}_{i y}$ we can find $\Delta t$

$\Delta \vec{d} = {\vec{v}}_{i y} \Delta t + \frac{\vec{a}}{2} \Delta {t}^{2}$

$0 = 12.56 \Delta t + \frac{- 9.8}{2} \Delta {t}^{2}$

The final solution is left for the reader. Note that some texts use -9.8 for gravity, some use -9.81, and some round to -10 for conceptual purposes, be sure to use the appropriate value for your class.