Question A.4: A ball is projected upwards at 16.0 m/s. Use the quadratic f...

From the discussion of ballistics in Chapter 2, we can write

x=\frac{1}{2} a t^{2}+v_{0} t+x_{0}                           (1)

The acceleration is due to gravity, given by a = -9.80 m/s²; the initial velocity is v_{0}=16.0 m / s ; and the initial position is the point of release, taken to be x_{0}=0 . Substitute these values into (1) and set x = 8.00 m, arriving at

x=-4.90 t^{2}+16.00 t=8.00

where units have been suppressed for mathematical clarity. Rearrange this expression into the standard form of Equation A.7:

a x^{2}+b x+c=0                        (A.7)

-4.90 t^{2}+16.00 t-8.00=0

The equation is quadratic in the time, t. We have a = -4.9, b = 16, and c = -8.00. Substitute these values into Equation A.8:

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}                    (A.8)

\begin{aligned}t &=\frac{-16.0 \pm \sqrt{16^{2}-4(-4.90)(-8.00)}}{2(-4.90)}=\frac{-16.0 \pm \sqrt{99.2}}{-9.80} \\&=1.63 \mp \frac{\sqrt{99.2}}{9.80}=0.614 s , 2.65 s\end{aligned}

Both solutions are valid in this case, one corresponding to reaching the point of interest on the way up and the other to reaching it on the way back down.