How High?- No Air Resistance
Suppose a small cannonball weighing 16 pounds is shot vertically upward, as shown in Figure 1, with an initial velocity v0 = 300 ft/s. The answer to the question “How high does the cannonball go?” depends on whether we take air resistance into account.
a) Suppose air resistance is ignored. If the positive direction is upward, then a model for the state of the cannonball is given by md2s/dt2 = -mg or d2s/dt2 = -g, where g is the force of the gravity. Take g = 32ft/s2, find the velocity v(t) of the cannonball at time t.
b) Use the result obtained in part (a) to determine the height s(t) of the cannonball measured from ground level. Find the maximum height attained by the cannonball.
If you plug in your numbers you will get a model of the form
h(t) = at^2 + bt + c
As with all such parabolas, the vertex is at (-b/2a, c - b^2/4a)
To find the velocity at time t (part a), we first need to solve the differential equation d^2s/dt^2 = -g. This equation represents Newton's second law for vertical motion, where g is the acceleration due to gravity (-32 ft/s^2 in this case).
To solve this differential equation, we can integrate it twice with respect to t. Let's go through the steps:
First, integrate d^2s/dt^2 = -g once with respect to t:
ds/dt = -gt + C1
Here, C1 is the constant of integration.
Next, integrate ds/dt = -gt + C1 once more with respect to t:
s(t) = -(1/2)gt^2 + C1t + C2
Here, C2 is the constant of integration.
Now, we have an equation describing the height s(t) of the cannonball as a function of time.
In part (b), we need to find the maximum height attained by the cannonball. To do this, we need to find the time t when the velocity v(t) is zero. Then, we can substitute this time into the equation for s(t) to find the height.
To determine when the velocity is zero, we set v(t) = ds/dt = -gt + C1 equal to zero and solve for t:
-gt + C1 = 0
gt = C1
t = C1/g
Now, we substitute t = C1/g into the equation s(t) = -(1/2)gt^2 + C1t + C2:
s(t) = -(1/2)g(C1/g)^2 + C1(C1/g) + C2
s(t) = -(C1^2)/(2g) + C1^2/g + C2
s(t) = (C1^2)/(2g) + C2
This gives us the height s(t) of the cannonball at time t. The maximum height attained by the cannonball is the value of s(t) at t = C1/g.