Question

A ball is launched straight up in the air from a height of 6 feet. Its velocity (feet/second) t seconds after launch is given by f(t)=-32t+285 Between 2 seconds and 9 seconds the ball's height changed by feet. Round answer to nearest tenth.)

Answer 1

Step-by-step explanation:

A ball is launched straight up in the air from a height of 6 feet. The velocity as a function of time t is given by :

f(t) = -32 t+285

Height of the ball is :

$h(t)=\int\limits{f(t){\cdot} dt}\\\\h(t)=\int\limits{(-32t+285){\cdot} dt}\\\\h(t)=-16t^2+285t+C$

C is constant. Here the ball is launched from a height of 6 feet. So,

$h(t)=-16t^2+285t+6$

At t = 2 s, $h(t)=-16(2)^2+285(2)+6=512\ m$

At t = 9 s, $h(t)=-16(9)^2+285(9)+6=1275\ m$

Between 2 s and 9 s, the ball's height changed is : 1275 - 512 = 763 m.