Question:

A football is kicked at ground level with a speed of 18.0 m/s at an angle of $31.0^\circ$ to the horizontal. How much later does it hit the ground?

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer:

$1.89\textrm{ s}$

### Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. To figure out the amount of time this ball spends in the air, we need to know what the initial vertical component of its velocity is. So knowing that it's kicked with a velocity of 18 meters per second at an angle of 31 degrees to the horizontal, we can figure out the vertical component of its velocity,*v y initial*by calculating

*v*times

*sin*of 31; we are using

*sin*because this is the opposite leg of this right triangle this velocity vector triangle. So that's 18 meters per second times

*sin*31 which gives 9.2707 meters per second upwards. And then we'll use this formula because we know that the displacement of this football is zero because it's gonna end at the same position that it started with the same vertical position that it started with so this vertical displacement will be zero. And then we can divide both of these both sides by

*t*, 0 divided by

*t*is 0 so it doesn't matter on the left and on the right-hand side, we get rid of this

*t*here and we get rid of one of the

*t*'s here and then we'll solve for this time here. We'll start by subtracting

*v y initial*over

*t*from both sides or sorry not over

*t*because the

*t*canceled right and then that's why it's negative in this next line. And so we have one-half acceleration in the

*y*-direction times time equals negative of the initial

*y*velocity and then we'll multiply by 2 over

*a y*in order to isolate

*t*. And so we have

*t*equals negative 2 times the initial

*y*velocity divided by the acceleration in the

*y*-direction which is negative 2 times 9.2707 meters per second divided by negative 9.8 which gives 1.89 seconds in the air.