Answer:
[tex]s(t)=s_{0}+v_{0}t-(0.5)(32)t^{2}[/tex]
Step-by-step explanation:
Let's use the definition of acceleration.
[tex] a(t)=dv/dt [/tex]
If we take the integral in both sides we will have:
[tex]\int\limits^t_{t_{0}} {a(t)} \, dt=\int\limits^v_{v_{0}} {dv}[/tex]
a(t) = -32, so it is independent of time.
[tex]a(t)\int\limits^t_{t_{0}} {dt}=\int\limits^v_{v_{0}} {dv}[/tex]
[tex]a(t)(t-t_{0})=v-v_{0}[/tex]
we can assume that [tex]t_{0} = 0[/tex]
[tex]v(t)=v_{0}+a(t)t[/tex] (1)
Using the definition of v(t) as the derivative of s (height) with t (time) we have:
[tex]v=ds/dt[/tex](2)
Taking the integral in both sides we can find s(t), and using (1) we have:
[tex]\int\limits^t_{t_{0}} {v(t)} \, dt=\int\limits^s_{s_{0}} {dx}[/tex]
Using (1) in (2)
[tex]\int\limits^t_{t_{0}}( v_{0}+ta(t))\, dt=\int\limits^s_{s_{0}} {ds}[/tex]
solving this integral, we have:
[tex] v_{0}t+0.5a(t)t^{2}=s(t)-s_{0}[/tex]
Finally, let's solve this equation for s(t).
[tex]s(t)=s_{0}+v_{0}t+0.5a(t)t^{2}[/tex]
[tex]s(t)=s_{0}+v_{0}t-(0.5)(32)t^{2}[/tex]
Have a nice day!
