The distance s (in m) above the ground for a projectile fired vertically upward with a velocity of 4[a] m/s as a function of time t (in s ) is given by s=4[a]t−4.[b]t2..a=3, b=7find LaTeX: t for LaTeX: v=0, find find LaTeX: v for LaTeX: t=4, find find LaTeX: v for LaTeX: t=5. What conclusions can you draw?Find the answers to these questions. What conclusions can you draw?

Answers:a) 1.224 sb) -27.2 m/sc) -37 m/sStep-by-step explanation:According to the given information the distance [tex]s[/tex] is given by:[tex]s=4(3)t-4(7)t^{2}=12 t - 28t^{2}[/tex] (1)In additio, since the projectile was fired vertically we can use the following equation:[tex]V=V_{o}-gt[/tex] (2)Where:[tex]V[/tex] is the final velocity[tex]V_{o}=12 m/s[/tex] is the initial velocity[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity[tex]t[/tex] is the timeKnowing this, let's begin:a) Find [tex]t[/tex]  for [tex]V=0[/tex]Using (2):[tex]0=V_{o}-gt[/tex] (3)[tex]0=12 m/s-9.8 m/s^{2}t[/tex] (4)[tex]t=\frac{12 m/s}{9.8 m/s^{2}}[/tex] (5)[tex]t=1.224 s[/tex] (6)b) Find [tex]V[/tex] for [tex]t=4 s[/tex]In this case we have to find [tex]V[/tex]:[tex]V=12 m/s-(9.8 m/s^{2})(4 s)[/tex] (7)[tex]V=-27.2 m/s[/tex] (8)c) Find [tex]V[/tex] for [tex]t=5 s[/tex][tex]V=12 m/s-(9.8 m/s^{2})(5 s)[/tex] (9)[tex]V=-37 m/s[/tex] (10)Conclusions:The time we found in a) [tex]t=1.224 s[/tex]when [tex]V=0[/tex] is at the maximum height of the projectile, just before going down. That is why in parts b) and c) when time is [tex]4 s[/tex] and [tex]5 s[/tex] the velocity is negative, because it is directed downwards.