# A projectile is shot at an angle of pi/6  and a velocity of  5 m/s. How far away will the projectile land?

Apr 24, 2017

2.2 m

#### Explanation:

To solve this kinematics problem, we have to consider motion in both the horizontal and vertical component. We then realise that the t value (time of flight) would be the same for both components.

Vertical component:

Given $u = 5 \sin \left(\frac{\pi}{6}\right) = 2.5 \frac{m}{s}$,
$v = - u = - 2.5 \frac{m}{s}$
and $a = - 9.81 \frac{m}{{s}^{2}}$ ,
find t.

$v = u + a t$
$t = \frac{v - u}{a} = \frac{- 2.5 - 2.5}{- 9.81} = 0.5097 s$

Horizontal component:

Given $t = 0.5097 s$
and $v = 5 \cos \left(\frac{\pi}{6}\right) = 4.330 \frac{m}{s}$ ,
find d.

distance
= speed x time
= (4.330)(0.5097)
= 2.207 m

Thus, the projectile will land 2.2 m away.