Answer:
[tex]\boxed{h(t)=-16t^2+24t+6}[/tex]
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:
[tex]h(t)=-16t^2+vt+h_{0}[/tex]
[tex]v[/tex] = initial vertical velocity of the ball in feet per second
[tex]h_{0}[/tex] = initial height of the ball in feet Complete the quadratic equation that models the situation.
From the graph we know:
[tex]h(0)=6 \\ \\ \therefore 6=-16(0)^2+v(0)+h_{0} \\ \\ \therefore h_{0}=6[/tex]
For a quadratic function:
[tex]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[/tex]
Finally:
[tex]\boxed{h(t)=-16t^2+24t+6}[/tex]
