Question

A soccer ball is kicked into the air from the ground. If the ball reaches a maximum height of 25 ft and spends a total of 2.5 s in the air, which equation models the height of the ball correctly? Assume that acceleration due to gravity is –16 ft/s2. h(t) = at2 + vt + h0

Answer 1

Answer:

Given the equation: $h(t) = at^2+vt+h_0$        .....[1]

If t = 0,  then h = 0.

Substitute these in [1] we get;

$h(0)=a \cdot (0)^2+v \cdot (0) + h_0$

$0=a \cdot (0)^2+v \cdot (0) + h_0$

Simplify:

$h_0 = 0$

Also, it is given that the acceleration due to gravity (a) = $-16ft/s^2$

then;

[1] ⇒ $h(t) = -16t^2+vt$

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;

$h(1.25) = -16(1.25)^2 +v(1.25)$

Solve for velocity(v);

$25= -16(1.25)^2 +v(1.25)$

$25 = 1.25(-16(1.25)+v)$

Divide both sides by 1.25 we get;

$20 =-16(1.25)+v$

$20 =-20+v$

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;

$h(t) = -16t^2+40t$  

Answer 2

Answer:

its A on ed-genuity

Step-by-step explanation: