Question

A baseball is thrown upward from ground level with an initial velocity of 48 feet per second, and its height h, in feet, is given by: h(t)=-16t^2+48t where t is time in seconds: Determine how many seconds after the baseball is thrown it reaches its maximum height. Then find the maximum height the baseball can reach. Need help from experienced tutors, thankyou

Answer 1

the vertex of any quadratic \(y=ax^2+bx+c\) has the first coordinate \(-\frac{b}{2a}\)
in your case \(a=-16,b=48\) and so \(-\frac{b}{2a}=-\frac{48}{2\times (-16)}=\frac{48}{32}=\frac{3}{2}\)

Answer 2

so one and a half seconds
since you know the time, you know the height by replacing \(t\) by \(\frac{3}{2}\) and see what you get

Answer 3

how would you find t

Answer 4

by plugging it in....right?

Answer 5

ok i got it thankyou