# A projectile is shot from the ground at a velocity of 4 m/s and at an angle of (5pi)/6. How long will it take for the projectile to land?

May 31, 2017

$0.4$ $\text{s}$

#### Explanation:

We're asked to find the time $t$ when the projectile lands with a known initial velocity and direction. In terms of physics, we need to find the time when the change in $y$ (the height) is equal to $0$.

We can use the equation

$\Delta y = {v}_{0 y} t - \frac{1}{2} g {t}^{2}$

to find the time $t$ when $\Delta y$ is zero,

where

• $\Delta y$ is the change in $y$-position (height) of the object, in $\text{m}$

• ${v}_{0 y}$ is the initial $y$-velocity of the object, in $\text{m"/"s}$

• $t$ is the time, in $\text{s}$, and

• $g$ is the acceleration due to gravity near Earth's surface, 9.80"m"/("s"^2).

To find the initial $y$-velocity, we can use the equation

${v}_{0 y} = {v}_{0} \sin \alpha$

where

• ${v}_{0}$ is the magnitude of the initial velocity, in $\text{m"/"s}$, and

• $\alpha$ is the initial launch angle.

Plugging in our known variables, we have

v_(0y) = v_0sinalpha = (4"m"/"s")sin((5pi)/6) = color(red)(2"m"/"s"

Now that we have our known variables necessary, let's plug them back into the original equation to find the times when the height is $0$ (there are two times because the height is zero when it is first launched as well):

$\Delta y = {v}_{0 y} t - \frac{1}{2} g {t}^{2}$

$0 = \left(\textcolor{red}{2 {\text{m"/"s"))t - 1/2(9.80"m"/("s}}^{2}}\right) {t}^{2}$

$\frac{1}{2} \left(9.80 \text{m"/("s"^2))t^2 = (2"m"/"s}\right) t$

We'll eliminate the $t = 0$ option by dividing both sides by $t$:

$\left(4.90 \text{m"/("s"^2))t = (2"m"/"s}\right)$

t = (2cancel("m")/cancel("s"))/(4.90cancel("m")/("s"^cancel(2))) = color(blue)(0.4"s"

rounded to one significant figure, which is the amount given in the problem.

Thus, the object will land again ad $t = 0.4 \text{s}$