the graph below shows the height of a projectile t seconds after it is launched. if acceleration due to…

Explanation:

Step1: Identify the value of (a)

The acceleration due to gravity (a=- 16), so the coefficient of (t^{2}) in the height - time formula (h(t)=at^{2}+vt + h_{0}) is (\frac{a}{2}=-8) since the general formula for vertical motion under constant acceleration (a) is (h(t)=\frac{a}{2}t^{2}+v_{0}t + h_{0}).

Step2: Use the initial - condition ((0,5))

Substitute (t = 0) and (h(0)=5) into (h(t)=at^{2}+vt + h_{0}). When (t = 0), (h(0)=a(0)^{2}+v(0)+h_{0}), so (h_{0}=5).

Step3: Use the vertex ((1,21))

The vertex of a parabola (y = ax^{2}+bx + c) (in our case (h(t)=at^{2}+vt + h_{0})) has (t=-\frac{v}{2a}). Since (a=-8) and (t = 1) at the vertex, we have (1=-\frac{v}{2\times(-8)}), which gives (1=\frac{v}{16}), so (v = 16).

Step4: Write the equation

Substitute (a=-8), (v = 16), and (h_{0}=5) into (h(t)=at^{2}+vt + h_{0}), we get (h(t)=-8t^{2}+16t + 5).

Answer:

(h(t)=-8t^{2}+16t + 5)