Suppose a small cannonball weighing 16 pounds is shot vertically upward with an initial
velocity Vo = 300 ft/s
.
a) Construct the mathematical model.
b) Suppose air resistance is ignored, determine the velocity of the cannonball at any time t.
c) Using the result obtained in part (a), determine the height s(t) of the cannonball
measured from the ground level. Determine the maximum height attained by the
cannonball.
d) Assume that air resistance is proportional to instantaneous velocity v(t). Show that in this
case the maximum height attained by the cannonball is less than that in part (b), by
supposing that the constant of proportionality is k = 0.0025.
v(t) = 300-32t
s(t) = 300t-16t^2
max s = 1406.25
(d) dv/dt = 300-32t-.0025v
v = 13100e^(-.0025t) - 12800
v=0 at x=9.267
s = 5240000(1-e^(-.0025t))-12800t
max s = 1384.7
How u got that V(t) = 300 - 32t
a) To construct the mathematical model, we need to consider Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the only force acting on the cannonball is gravity, so we can use the equation:
F = mg
Where F is the force of gravity, m is the mass of the cannonball, and g is the acceleration due to gravity (approximately 32.2 ft/s^2).
Given that the weight of the cannonball is 16 pounds (lb), we can convert it to mass using the conversion factor 1 lb = 0.03125 slugs:
m = 16 lb * 0.03125 slug/lb = 0.5 slug
Therefore, the force of gravity is:
F = 0.5 slug * 32.2 ft/s^2 = 16.1 ft * slug / s^2
b) To determine the velocity of the cannonball at any time t, we can use the equation for acceleration:
F = ma
Rearranging the equation, we have:
a = F / m
Since the only force acting on the cannonball is gravity, and acceleration is the derivative of velocity, we can write:
dv/dt = F / m
We can substitute the formula for the force of gravity into this equation:
dv/dt = (16.1 ft * slug / s^2) / 0.5 slug
dv/dt = 32.2 ft/s^2
Integrating both sides of the equation with respect to time (t), we get:
∫ dv = ∫ 32.2 ft/s^2 dt
v = 32.2 ft/s^2 * t + C
where C is the constant of integration.
To find the initial velocity (Vo), we can substitute the given values into the equation:
300 ft/s = 32.2 ft/s^2 * 0 + C
C = 300 ft/s
Therefore, the equation for velocity as a function of time is:
v(t) = 32.2 ft/s^2 * t + 300 ft/s
c) To determine the height (s) of the cannonball measured from the ground level, we can integrate the equation for velocity with respect to time:
ds/dt = v(t)
Integrating both sides of the equation:
∫ ds = ∫ (32.2 ft/s^2 * t + 300 ft/s) dt
s = (16.1 ft/s^2 * t^2) / 2 + 300 ft/s * t + D
where D is the constant of integration.
To find the initial height (s0), we can substitute the given values into the equation:
s(0) = (16.1 ft/s^2 * (0^2)) / 2 + 300 ft/s * 0 + D
s0 = D
Therefore, the equation for height as a function of time is:
s(t) = (16.1 ft/s^2 * t^2) / 2 + 300 ft/s * t + s0
To find the maximum height attained by the cannonball, we need to find the time (t) at which the velocity (v) becomes zero. By setting v(t) = 0 and solving for t:
0 = 32.2 ft/s^2 * t + 300 ft/s
-32.2 ft/s^2 * t = 300 ft/s
t = 300 ft/s / -32.2 ft/s^2
t ≈ -9.315 s (Ignoring the negative value since it's not physically meaningful)
Substituting this time back into the equation for height:
s(t) = (16.1 ft/s^2 * (-9.315 s)^2) / 2 + 300 ft/s * (-9.315 s) + s0
s(t) ≈ -441.81 ft + s0
Therefore, the maximum height attained by the cannonball is -441.81 ft above the initial height (s0), which is below the ground level.
d) If air resistance is proportional to instantaneous velocity (v), we can express it as:
F = -kv
Where F is the force due to air resistance and k is the constant of proportionality (k = 0.0025).
Applying Newton's second law, the net force acting on the cannonball is:
F = ma
Including the force due to air resistance, the equation becomes:
F = mg - kv
Hence, the equation for acceleration becomes:
dv/dt = (mg - kv) / m
Substituting the weight of the cannonball (16.1 ft * slug / s^2) and mass (0.5 slug):
dv/dt = (16.1 ft/s^2 - 0.0025 * v) / 0.5
Simplifying the equation:
2dv / (16.1 ft/s^2 - 0.0025 * v) = dt
Solving this separable first-order differential equation will result in the equation for velocity as a function of time. By integrating both sides of the equation, we obtain:
2∫ (16.1 ft/s^2 - 0.0025 * v) dv = ∫ dt
16.1v - 0.0025v^2 / 2 = t + C
Multiplying through by 2 to eliminate the fraction:
32.2v - 0.00125v^2 = 2t + 2C
Rearranging the equation:
-0.00125v^2 + 32.2v - 2t = 2C
To find the maximum height attained by the cannonball in this case, we need to determine the vertex of the graph formed by the equation above. The maximum height corresponds to the highest point on the graph, which occurs when the velocity is zero.
Setting v = 0 and solving for t:
-0.00125(0)^2 + 32.2(0) - 2t = 2C
-2t = 2C
t = C
Therefore, the maximum height attained by the cannonball in this case is C units above the initial height. Since C is a constant, it will be the same as in part (c).
a) To construct the mathematical model, we need to analyze the motion of the cannonball and apply the laws of physics. In this case, we can use Newton's laws of motion and the equations of motion to describe the vertical motion of the cannonball.
b) To determine the velocity of the cannonball at any time t, we can use the equation of motion for velocity. Since the cannonball is shot vertically upward, the only force acting on it is the force due to gravity. Therefore, we can use the formula:
v(t) = Vo - gt
where:
- v(t) represents the velocity of the cannonball at time t.
- Vo is the initial velocity of the cannonball.
- g is the acceleration due to gravity (approximately 32.2 ft/s^2).
By substituting the given values, we can calculate the velocity of the cannonball at any time t.
c) To determine the height s(t) of the cannonball measured from the ground level, we need to integrate the equation of motion for velocity with respect to time, which gives us the equation of motion for height. Considering that the initial height of the cannonball is 0, we have:
s(t) = ∫(Vo - gt) dt
Integrating the equation and substituting the given initial velocity and acceleration due to gravity, we can calculate the height of the cannonball at any time t.
To find the maximum height attained by the cannonball, we need to determine the time when the velocity of the cannonball becomes zero (reaches its peak) and plug that value into the equation for height.
d) In this case, where air resistance is proportional to instantaneous velocity v(t), the equation for the velocity of the cannonball can be modified as:
v(t) = Vo - (g + kv(t))t
where:
- k is the constant of proportionality for air resistance (given as 0.0025).
Using the same method as in part (c), by integrating the modified equation of motion for velocity, we can calculate the height of the cannonball at any time t.
Comparing the two height calculations, we can determine that the maximum height attained by the cannonball in the presence of air resistance is less than the maximum height without air resistance (as given in part b).