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?

1 Answer
Apr 21, 2018

"Elapsed Time"=2.74s

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*sin(theta)

V_v=19sqrt2/2approx13.4m/s

Note that by launching a projectile at the angle (3pi)/4 you are effectively shooting it backwards at an angle of 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_(vf)=V_(vi)-at

Here, the final vertical velocity V_(vf), is zero and acceleration due to gravity on earth is -9.81 m/s^2

0=13.4-9.81t

t=13.4/9.81approx1.37s

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

"Elapsed Time"=2.74s