Question

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

Answer 1

24

Explanation:

cause

Answer 2

`\boxed{h(t)=-16t^2+24t+6}`

Step-by-step explanation:

From the statement of the problem we know:

The graph shows the height (h), in feet, of a basketball t seconds after it is shot. Projectile motion formula:

`h(t)=-16t^2+vt+h_(0)`

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

`h_(0)` = initial height of the ball in feet Complete the quadratic equation that models the situation.

From the graph we know:

`h(0)=6 \\ \\ \therefore 6=-16(0)^2+v(0)+h_(0) \\ \\ \therefore h_(0)=6`

For a quadratic function:

`f(x)=ax^2+bx+c \\ \\ \\ The \ vertex \ is: \\ \\ V(-(b)/(2a),f(-(b)/(2a))) \\ \\ Since: \\ \\ h(t)=-16t^2+vt+6 \ then: \\ \\ a=-16 \\ b=v \\ c=h_(0)=6 \\ \\ So: \\ \\ -(b)/(2a)=-(v)/(2(-16))=0.75 \\ \\ (v)/(32)=0.75 \\ \\ \therefore v=32(0.75) \ \therefore v=24`

Finally:

`\boxed{h(t)=-16t^2+24t+6}`