Question

an object is launched at 25 m/s from a height of 80 m. the equation for the height (h) in terms of time (t) is given by h(t)=4.9t^2+25t+80. what's the objects maximum height?

Answer 1

There must be minus sign with 4.9 .

$h(t) = -4.9t^2+25t +80$

The given equation is the equation of parabola , which is maximum at its vertex, and the formula of vertex is

$t=-\frac{b}{2a}=-\frac{25}{2(-4.9)} = 2.55 seconds$

So the height is maximum at t=2.55 seconds and to find the maximum height , we need to plug 2.55 for t , that is

$h(2.55)=-4.9(2.55)^2+25(2.55)+80 = Approx \ 112 \ m$

So the maximum height is 112 m .