# A projectile is shot from the ground at a velocity of 1 ms^-1 at an angle of pi/3. How long will it take for the projectile to land?

Jan 15, 2016

The total time taken for the projectile to return to earth will be 0.176 s.

#### Explanation:

The vertical component of the initial velocity, ${v}_{y}$, is given by ${v}_{y} = v \sin \theta = 1 \sin \left(\frac{\pi}{3}\right) = 0.866 m {s}^{-} 1$

At the top of its path the vertical component of the velocity will be $0$.

Use the formula $v = u + a t$ where:

• $v$ is the final velocity $\left(m {s}^{-} 1\right)$, $0 m {s}^{-} 1$,
• $u$ is the initial velocity $\left(m {s}^{-} 1\right)$, $0.866 m {s}^{-} 1$,
• $a$ is the acceleration $\left(m {s}^{-} 2\right)$, which in this case is the acceleration due to gravity, $- 9.8 m {s}^{-} 2$ (negative because it is in
the opposite direction to the initial velocity, and
• $t$ is the time $\left(s\right)$.

Rearranging to make the time the subject:

$t = \frac{v - u}{a} = \frac{0 - 0.866}{-} 9.8 = 0.088 s$

This is the time from launch to the top of the trajectory, and the projectile will take the same amount of time to return to earth, so this time should be doubled for the whole motion, so $t = 0.176 s$.

This makes sense as an answer due to the very low initial velocity.