# If a projectile is shot at a velocity of 18 m/s and an angle of pi/3, how far will the projectile travel before landing?

##### 1 Answer

Horizontal distance $x = 28.63186029 \text{ }$meters

#### Explanation:

Let $t$ be the time required for the projectile to reach the maximum height.

Let $2 t$ be the time required for the projectile to travel from initial position to the final position at landing.

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

$t = \frac{- 18 \cdot \sin {60}^{\circ}}{- 9.8}$

$t = \frac{45 \sqrt{3}}{49} = 1.590658905 \text{ }$sec

For horizontal distance $x$

$x = {v}_{0} \cdot \cos \theta \cdot 2 t$

$x = 18 \cdot \cos {60}^{\circ} \cdot 2 \left(1.590658905\right)$

$x = 28.63186029 \text{ }$meters

God bless....I hope the explanation is useful.