Answer:
Given the equation: [tex]h(t) = at^2+vt+h_0[/tex] .....[1]
If t = 0, then h = 0.
Substitute these in [1] we get;
[tex]h(0)=a \cdot (0)^2+v \cdot (0) + h_0[/tex]
[tex]0=a \cdot (0)^2+v \cdot (0) + h_0[/tex]
Simplify:
[tex]h_0 = 0[/tex]
Also, it is given that the acceleration due to gravity (a) = [tex]-16ft/s^2[/tex]
then;
[1] ⇒ [tex]h(t) = -16t^2+vt[/tex]
If the ball reaches a maximum height of 25 ft and spends a total of 2.5 s in the air.
⇒ total time = 2.5 s
The height of ball is maximum at t =1.25 s
⇒h(1.25) = 25 ft
we have;
[tex]h(1.25) = -16(1.25)^2 +v(1.25)[/tex]
Solve for velocity(v);
[tex]25= -16(1.25)^2 +v(1.25)[/tex]
[tex]25 = 1.25(-16(1.25)+v)[/tex]
Divide both sides by 1.25 we get;
[tex]20 =-16(1.25)+v[/tex]
[tex]20 =-20+v[/tex]
Add 20 both sides to an equation we get;
v = 40 ft\s
Therefore, the equation become to models the height of the ball is;
[tex]h(t) = -16t^2+40t[/tex]
