Question

A ball is launched upward at 48 ft/s from a platform that is 100 ft. high. This can be modeled by the equation

Answer 1

Answer:

1.5 seconds.

Step-by-step explanation:

The height of ball modeled by the equation:

$h(t)=-16t^2+48t+100$

where, t is time in seconds.

The leading coefficient is negative, it means the given function is a downward parabola and vertex of the downward parabola is the point of maxima.

If a parabola is defined by $f(x)=ax^2+bx+c$, then the vertex of the parabola is

$\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

In the given equation, a=-16, b=48 and c=100.

$\dfrac{-b}{2a}=\dfrac{-48}{2(-16)}=1.5$

$f(\dfrac{-b}{2a})=f(1.5)=-16(1.5)^2+48(1.5)+100=136$

Vertex of the given parabola is (1.5,136).

Therefore, the maximum height of ball is 136 ft after 1.5 seconds.