3.12 . A rookie quarterback throws a football with an initial upward velocity component of 12.0 m/s and a horizontal velocity component of 20.0 m/s. Ignore air resistance. (a) How much time is required for the football to reach the highest point of the trajectory? (b) How high is this point? (c) How much time (after it is thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)? (d) How far has the football traveled horizontally during this time? (e) Draw x-t, y-t, and graphs for the motion.
SOLUTION
From the figure, -y direction is downward. Consider the following
$v_{0x}=20.0 m/s $
$v_{0y}=12.0 m/s $
$a_x=0$, and $a_y=-g=-9.8 m/s^2$
at the highest point of the trajectory, $v_y=0$
a) Solving for the time,
$v_y=v_{0y}+a_yt$
$0=12.0 m/s + a_yt$
$-12.0 m/s = a_yt$
$\frac{-12.0 m/s}{-g} = t$
$t=1.22s$
b) We may choose from the following equations, but the answer is the same:
$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$
$v^2_y=v^2_{0y}+2a_y (y-y_0)$
$y=y_0+(\frac{v_{0y}+v_y}{2})t
From $y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$:
$y=\frac{1}{2}a_yt^2=-\frac{1}{2}gt^2$
$y=-\frac{1}{2}(9.8 / m/s^2)(1.22s)^2=7.3 / m$
From $v^2_y=v^2_{0y}+2a_y(y-y_0)$
$0= v^2_{0y}-2gy$
$y=\frac{ v^2_{0y}}{2g}=7.3 / m$
From $y=y_0+(\frac{v_{0y}+v_y}{2})t
$y= 0+(\frac{12.0 / m/s / + 0}{2})(1.22s)=7.3 / m$
