Billy band hit a fly ball. The height in feet,h, of the ball abive the ground t seconds after he hit it can be found using the equation h=-16t+96t+4

Question

How high above the ground was the ball the moment billy hit it ?

to the nearest second, how long after billy hit the ball did it reach its greatest height ?

to the nearest second , how long after billy hit the ball did it hit the ground?

to the nearest foot , what is the greatest height the ball reached

Answer 1

The equation is h = -16*t^2 + 96 t + 4

The moment Billy hit the ball at time t=0, the ball is 4 feet above the ground, as -16*0^2 + 96(0) +4 = 4

Next the quadratic function has coefficients A=-16, B=96 and C=4.

The extrema occurs at x=-B/(2a) = -96/(2*-16) = -96/-32 = 3

So at t=3, -16*3^2 + 96*3 + 4 = -144 + 288 + 4 = 147 feet

The ball hits the ground at 0 = -16t^2 + 96t + 4

0 = 4t^2 - 24t - 1 <--- divides by -4

Quadratic formula says:

[24 + sqrt( (-24)^2 + 4*4*1)] / 8 <--- A=4, B=-24, C=-1; subtracting 2 negatives makes the sign +

[24 + sqrt( 576 + 16)]/8

[24 + sqrt(592)] / 8 = 6.0413812....

So to the nearest second, it takes 6 seconds for the ball to hit the ground