# A projectile is shot at an angle of (5pi)/12  and a velocity of  5 m/s. How far away will the projectile land?

May 24, 2017

1.275 m

#### Explanation:

First, break the initial velocity of the object into $x$ and $y$ components.

${v}_{x} = 5 \cos \left(\frac{5 \pi}{12}\right) \frac{m}{s} = 1.294 \frac{m}{s}$

${v}_{y} = 5 \sin \left(\frac{5 \pi}{12}\right) \frac{m}{s} = 4.830 \frac{m}{s}$

The formula for figuring out how far a projectile will fly is:

$d = \text{fall time} \times {v}_{x}$

The fall time can be found using the kinematics equation:

$y = {y}_{0} + {v}_{y} t + \frac{1}{2} a {t}^{2}$

In this case, we know ${y}_{0} = 0$ ${v}_{y} = 4.830$ and $a = g = - 9.8 \frac{m}{s} ^ 2$, and we're trying to find the time when $y = 0$ (when it hits the ground again).

$0 = 4.830 t - 4.9 {t}^{2}$

$0 = t \left(4.830 - 4.9 t\right)$

$t = 0 \mathmr{and} 4.9 t = 4.830$

$t = 0 \mathmr{and} t = 0.9857$

So the object has a fall time of $0.9857$ seconds.

Now all we have to do is use the projectile distance formula:

$d = \text{fall time} \times {v}_{x}$

$d = 0.9857 \cdot 1.294 = 1.275 m$

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As a side note for the future, when the starting height of an object is $0$, you can derive a formula for projectile distance which will save time:

$y = {y}_{0} + {v}_{y} t + \frac{1}{2} a {t}^{2}$

$0 = v t \sin \theta - \frac{g}{2} {t}^{2}$

$0 = t \left(v \sin \theta - \frac{g}{2} t\right)$

$0 = v \sin \theta - \frac{g}{2} t$

$\frac{g}{2} t = v \sin \theta$

$t = \frac{2 v \sin \theta}{g}$

Plugging in this value for the fall time, we get:

$d = \text{fall time} \times {v}_{x}$

$d = \frac{2 v \sin \theta}{g} \times v \cos \theta$

$d = \frac{{v}^{2} \left(2 \sin \theta \cos \theta\right)}{g}$

$d = \frac{{v}^{2} \sin \left(2 \theta\right)}{g}$