The height \(x\) above the ground of a vertically launched projectile is given by \(x(t)=p t-q t^{2}\), with \(p=42 \mathrm{~m} / \mathrm{s}\) and \(q=4.9 \mathrm{~m} / \mathrm{s}^{2}\).

(a) At what instant is the projectile at a height of \(20 \mathrm{~m}\) ?

(b) What is the meaning of the two solutions obtained in part \(a\) ?

(c) Sketch a graph of the \(x\) component of the projectile's velocity as a function of time.