1. ## football hangtime.

could i use trig with this? to find the x0? then use one of the common physics 'formulas to find time?

A punter on a football team tries to kick a football so that it stays in the air for a long "hang time."
the ball is kicked with an initial velocity of 25.0 m/s at an angle of 58.5° above the ground, what is the "hang time"?
s

2. Originally Posted by rcmango
could i use trig with this? to find the x0? then use one of the common physics 'formulas to find time?

A punter on a football team tries to kick a football so that it stays in the air for a long "hang time."
the ball is kicked with an initial velocity of 25.0 m/s at an angle of 58.5° above the ground, what is the "hang time"?
s
It's not so much an equation as a method.

Set up the following coordinate system:
Place an origin where the football starts and let the +x direction be in the horizontal direction of the ball and +y be straight up.

Presuming the football starts on the ground and ends on the ground we have the following information:
$\displaystyle x_0 = 0~m$ and $\displaystyle y_0 = 0~m$
$\displaystyle x = ?$ and $\displaystyle y = 0~m$
$\displaystyle v_{0x} = 25.0 \cdot cos(58.5^o)~m/s$ and $\displaystyle v_{0y} = 25.0 \cdot sin(58.5^o)~m/s$
$\displaystyle a_x = 0~m/s^2$ and $\displaystyle a_y = -9.8~m/s^2$

We want an equation for t. It looks like
$\displaystyle y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$
will do the trick:

$\displaystyle 0 = 0 + 25 \cdot sin(58.5^o)~t - 4.9t^2$ <-- Divide through by t

$\displaystyle 0 = 25 \cdot sin(58.5^o) - 4.9t$

$\displaystyle t = \frac{25~sin(58.5^o)}{4.9} = 4.73707~s$

-Dan