# A projectile is shot from the ground at a velocity of 19 m/s and at an angle of (3pi)/4. How long will it take for the projectile to land?

Apr 21, 2018

$\text{Elapsed Time} = 2.74 s$

#### Explanation:

The best thing to do when asked for time is to focus only on the vertical components of your velocity since it will be because of them that your projectile ends up on the ground.

First lets find what the vertical component of your projectile's velocity is. If this is a jump, I would recommend drawing a right triangle and solving it that way but for the sake of simplicity, the vertical velocity is

${V}_{v} = 19 \cdot \sin \left(\theta\right)$

${V}_{v} = 19 \frac{\sqrt{2}}{2} \approx 13.4 \frac{m}{s}$

Note that by launching a projectile at the angle $\frac{3 \pi}{4}$ you are effectively shooting it backwards at an angle of $\frac{\pi}{4}$ to the horizontal.

To find the time it takes for the projectile to rise and fall, it is only necessary to find the time it takes for the projectile to rise and then double this quantity.

To find the time it takes to rise, use the equation

${V}_{v f} = {V}_{v i} - a t$

Here, the final vertical velocity ${V}_{v f}$, is zero and acceleration due to gravity on earth is $- 9.81 \frac{m}{s} ^ 2$

$0 = 13.4 - 9.81 t$

$t = \frac{13.4}{9.81} \approx 1.37 s$

To find the total elapsed time, remember you must double this.

$\text{Elapsed Time} = 2.74 s$