Answer:
a) v(t)=-32t+300
b) s(t)=-16t^{2}+300t
[tex]s_{max} =\frac{5625}{4}ft=1406.25 ft[/tex]
Step-by-step explanation:
A good way to start solving this problem out is to draw a diagram of the situation (see attached picture).
a)
For part a, we need to find an equation for the velocity of the cannon ball. We have an equation of its acceleration:
[tex]a=\frac{dv}{dt}=\frac{d^{2}s}{dt^{2}}=-g[/tex]
we know that:
[tex]g=32 ft/s^{2}[/tex]
so we got the following differential equation:
[tex]\frac{dv}{dt} =-32[/tex]
we can find the function for velocity by integrating, so we get:
[tex]v =\int\limits {-32} \, dt[/tex]
which yields:
v(t)=-32t+C
we need to find the value of C. The problem tells us that v(0)=300 ft/s so we can use this to find c:
300=-32(0)+C
so
C=300
therefore the equation for velocity is:
v(t)=-32t+300
b)
For part b) we need to find an equation for the height of the cannonball. We know that:
[tex]v=\frac{ds}{dt}[/tex]
so:
[tex]\frac{ds}{dt}=-32t+300[/tex]
so we can integrate this equation to find the function for height of the ball, so we get:
[tex]s=\int\limits {-32t+300} \, dt[/tex]
which yields:
[tex]s(t)=-\frac{32t^{2}}{2}+300t+C[/tex]
which can be simplified to:
[tex]s(t) =-16t^{2}+300t+C[/tex]
We know the height is measured from the ground level, this means that s(0)=0 so we can find C with this information:
[tex]0=-16(0)^{2}+300(0)+C[/tex]
so C=0, therefore our equation is:
[tex]s(t)=-16t^{2}+300t[/tex]
in order to find the maximum height attained by the cannonball, we can set the velocity function equal to zero and solve for t, so we get:
0=-32t+300
-32t=-300
[tex]t=\frac{300}{32}[/tex]
[tex]t=\frac{75}{8} s[/tex]
and substitute it into our position function so we get:
[tex]s(\frac{75}{8})=-16(\frac{75}{8})^{2}+300(\frac{75}{8})[/tex]
which yields:
[tex]s_{max} =\frac{5625}{4}ft=1406.25 ft[/tex]
