# A projectile is shot from the ground at a velocity of 4 m/s at an angle of pi/3. How long will it take for the projectile to land?

Jan 5, 2016

$0.7 \text{s}$

#### Explanation:

Use the equation of motion:

$v = u + a t$

The vertical component of the initial velocity is $v \sin \theta$.

At the maximum height the final velocity is zero. We can find the time taken to reach this point.

At this point the equation becomes:

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

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

$\pi = {180}^{\circ}$
$\therefore \frac{\pi}{3} = {180}^{\circ} / 3 = {60}^{\circ}$
$\therefore t = 4 \sin \frac{60}{9.8} = 0.35 \text{s}$
The second leg of the flight is when the projectile falls to earth. Because the path is symmetrical this will also be equal to $0.35 \text{s}$.
So total time of flight = $0.35 \times 2 = 0.7 \text{s}$