Homework question

Question

A juggler tosses a ball into the air. The ball leaves the juggler’s hand 4 feet above the ground and has an initial vertical velocity of 30 feet per second.

This can be modeled by the equation, `h(t)=-16t^{2}+30t+4`

Suppose, the juggler catches the ball when it falls back to a height of 3 feet. How long is the ball in the air?

Which method would be best in order to solve this?

Quadratic Formula

Graphing

Isolating the Variable

Answer 1

h(t) = -16t² + 30t + 4 = 3; so, we get quadratic equatin: 16t² - 30t - 1 = 0.

I prefer TI-84 Zero command from Calc Menu: t = 1.9 s

Answer 2

h(0) = 4, with t=0

h(t) = 3 = -16t^2 + 30t + 4, solve for t

-16t^2 + 30t = -1

t = -30/-32 + or - (1/-32)sqr(900+64)

= 15/16 + 0.97 = .97+.94 = 1.91 seconds from 4 to 3 feet

it reaches a maximum height at time t when -32t +30 = 0

t =15/16 = 0.97 seconds. .94 seconds later, it reaches 3 feet in height, one foot lower than it started

quadratic formula is best method

graphing the parabola (the quadratic equation) is helpful to visualize the path of the ball up and down