A projectile is shot from the ground at a velocity of 12 m/s and at an angle of (7pi)/12. How long will it take for the projectile to land?

Jun 3, 2017

$2.4$ $\text{s}$

Explanation:

We're asked to find the time when the projectile lands when it is launched with a known initial velocity. We need to find the time $t$ when the height $\Delta y$ is $0$.

We can use the equation

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

to find this. Our known quantities are

• $\Delta y$ must be $0$, because we're trying to find the time when this occurs.

• to find the initial $y$-velocity ${v}_{0 y}$, we can use the equation ${v}_{0 y} = {v}_{y} \sin \alpha$, so ${v}_{0 y} = 12 \text{m"/"s"sin((7pi)/12) = 11.6"m"/"s}$

• $g$, the acceleration due to gravity near Earth's surface, is 9.8"m"/("s"^2)

If we make $\Delta y$ zero, we can rearrange the equation to solve for $t$:

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

$t = \frac{2 {v}_{0 y}}{g}$

Therefore,

t = (2(11.6cancel("m")/"s"))/(9.8cancel("m")/(cancel("s"^2))) = color(red)(2.4 color(red)("s"

The projectile will land after $2.4$ seconds.