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

Answer:

h(t) = –16t2 + 24t + 6

Answer 2

Answer:

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

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:

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

 Finally:

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