The graph shows the height (h), in feet, of a basketball t seconds after it is shot. projectile motion formula: h(t) = -16t2 + vt + h0 v = initial vertical velocity of the ball in feet per second h0 = initial height of the ball in feet complete the quadratic equation that models the situation. h(t) = –16t2 + t + 6 image description

h(t) = –16t^2 + 46*t + 6

Step-by-step explanation:

To quickly solve this problem, we can use a graphing tool or a calculator to plot the equation.

Please see the attached image below, to find more information about the graph

The projectile formula is

h(t) = -16*t^2 + v*t + h0

Where,

v = initial vertical velocity of the ball in feet per second

h0 = initial height of the ball in feet

We need to fill the equation

h(t) = –16t^2 + v*t + 6

Where

v is the missing speed

If we go through many values of v until we find the same graph as the one showed in the first picture, we can find that

v = 46 ft/s

a 270° rotation about point p says on e d g e n u it y (had to space it out because it says it's a misspelling so i couldn't submit it

1/2*qr*rs

step-by-step explanation: