# A projectile is shot from the ground at a velocity of 4 m/s at an angle of pi/6. How long will it take for the projectile to land?

Mar 3, 2016

$4 \cdot \frac{\sqrt{3}}{5} m$

#### Explanation:

Let's start with Newton's equation $\ast F \ast = m \cdot \ast a \ast$. As the problem is bi-dimensional, the previous equation can be split in the horizontal and vertical components

${F}_{x} = m \cdot {a}_{x}$ for the horizontal component and

${F}_{y} = m \cdot {a}_{y}$ for the vertical component.

The horizontal acceleration is null while the vertical acceleration is equal to gravity

${a}_{x} = 0$ ; ${a}_{y} = - g$

and using the definition of acceleration

a_x = (d^2x)/dt^2 ; a_y = (d^2y)/dt^2

and integrating both part of equations

x= v_(x0)*t ; y= v_(yo)*t - g*t^2/2 (1)

The projectile will land when $y = 0 \implies$ introducing it in (1)

0= v_(yo)*t_l - g*t_l^2/2 ; t_l = 2*v_(yo)/g

and replacing in the equation or $x$

${x}_{l} = 2 \cdot {v}_{x 0} \cdot {v}_{y 0} / g$

The components of shot speed are

${v}_{x 0} = {v}_{0} \cdot \cos \left(\frac{\pi}{6}\right)$

${v}_{y 0} = {v}_{0} \cdot \sin \left(\frac{\pi}{6}\right)$

thus

${x}_{l} = 2 \cdot {v}_{0}^{2} / g \cdot \cos \left(\frac{\pi}{6}\right) \cdot \sin \left(\frac{\pi}{6}\right)$

and replacing by the figures given in the problem

${x}_{l} = 2 \cdot \frac{16}{10} m \cdot \cos \left(\frac{\pi}{6}\right) \cdot \sin \left(\frac{\pi}{6}\right) = \frac{16}{5} m \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} =$ $= 4 \cdot \frac{\sqrt{3}}{5} m$