Difficulty: Intermediate Prerequisites: Integration of vector functions, trigonometric decomposition of velocity, basic kinematics (distance, velocity, acceleration).
Section 13.4 applies the vector calculus you have been building to a classic physics scenario: projectile motion. Instead of treating horizontal and vertical components separately (as in physics courses), you work with a single vector equation of motion. You start from an acceleration vector, integrate once to get velocity, integrate again to get position, and pin down the constants of integration using initial conditions. This vector approach handles extra forces (like wind) cleanly, because they just add another component to the acceleration vector.
To solve a projectile problem with vectors, write a(t) as a constant vector, integrate twice to get v(t) and s(t), use initial conditions to find the integration constants, then set the height component to zero to find the landing time and position. Wind or drag simply changes the acceleration vector.
Acceleration vector a(t)
The second derivative of the position vector. In projectile problems with constant forces (gravity, wind), a(t) is a constant vector. Think of it as the list of forces acting on the object, divided by mass, written component by component.
Velocity vector v(t)
The first derivative of the position vector, obtained by integrating a(t). It tells you the speed and direction of the object at time t. Each component picks up a constant of integration, determined by the initial velocity.
Position vector s(t)
Obtained by integrating v(t). It gives the location of the object at time t. Another set of constants of integration appears, determined by the initial position.
Initial conditions
The known values of velocity and position at t = 0. These fix the constants of integration. For a projectile launched from the origin at angle theta with speed v_0, the initial velocity is
Landing time
The positive value of t at which the height (y-component of position) returns to zero. Found by setting the y-component of s(t) equal to zero and solving for t.
Every projectile problem in this section follows the same four steps:
Write the acceleration vector a(t) from the given forces (gravity downward, wind horizontal, etc.)
Integrate a(t) to get v(t), introducing constants C_1, C_2
Apply the initial velocity condition v(0) =
Integrate v(t) to get s(t), introduce new constants C_3, C_4, and apply the initial position s(0) to find them
Once you have s(t), set the y-component equal to zero to find the landing time, then plug that time back into the x-component to find the horizontal distance.
A cannon fires at 30 degrees above the horizontal with initial speed 747 ft/s. Gravity acts downward at 32 ft/s^2. Wind pushes horizontally against the motion at 0.5 ft/s^2. Initial height is 0 ft.
Step 1: acceleration
a(t) = <-0.5, -32>
The negative signs reflect that wind opposes horizontal motion and gravity pulls downward
Step 2: integrate for velocity
v(t) = <-0.5t + C_1, -32t + C_2>
Step 3: apply initial velocity
v(0) = <747 cos(30), 747 sin(30)> = <747 sqrt(3)/2, 747/2>
So C_1 = 747 sqrt(3)/2 and C_2 = 747/2
v(t) = <-0.5t + 747 sqrt(3)/2, -32t + 747/2>
Step 4: integrate for position
s(t) = <-0.25t^2 + (747 sqrt(3)/2) t + C_3, -16t^2 + (747/2) t + C_4>
s(0) = <0, 0>, so C_3 = 0 and C_4 = 0
s(t) = <-0.25t^2 + (747 sqrt(3)/2) t, -16t^2 + (747/2) t>
Finding the landing point
Set y-component to zero: -16t^2 + (747/2) t = 0
Factor: t(-16t + 747/2) = 0
Solutions: t = 0 (launch) or t = 747/32 (landing)
Substitute t = 747/32 into the x-component to get horizontal distance
The numerical answer is left to a calculator
The key takeaway: wind changes only the x-component of acceleration. The y-component (and therefore the time of flight) is unaffected by horizontal wind in this model.
\vec{a}(t) = \langle a_x,\, -g \rangle \quad \text{(constant acceleration with horizontal force and gravity)}\vec{v}(t) = \int \vec{a}(t)\, dt = \langle a_x t + C_1,\, -gt + C_2 \rangle\vec{s}(t) = \int \vec{v}(t)\, dt = \left\langle \tfrac{1}{2}a_x t^2 + C_1 t + C_3,\, -\tfrac{1}{2}g t^2 + C_2 t + C_4 \right\rangle\vec{v}(0) = \langle v_0 \cos\theta,\, v_0 \sin\theta \rangle \quad \text{(initial velocity from launch angle)}t_{\text{land}} = \frac{v_0 \sin\theta}{\tfrac{1}{2}g} = \frac{2 v_0 \sin\theta}{g} \quad \text{(landing time, no wind, launched from ground level)}When wind or drag is present, the x-acceleration is no longer zero, but the structure of the solution is identical: integrate, apply initial conditions, solve.
This is the mathematics behind ballistics, sports science (predicting where a ball will land), and aerospace trajectory planning. The vector approach extends naturally to three dimensions, which is how actual flight paths are computed when crosswinds or lateral forces are present.
Students often forget that integrating a vector function means integrating each component separately. Each component gets its own constant of integration.
A frequent error is using the wrong sign on gravity. If the y-axis points upward, gravitational acceleration is negative (pointing down).
Students sometimes apply the initial velocity magnitude directly as a component, forgetting to decompose it into cos and sin components using the launch angle.
Confusing the time the projectile lands (when y = 0 again) with the time it reaches maximum height (when v_y = 0). These are different questions.
This topic combines integration, initial-value problems, and trigonometry in one question. It tests a broad range of skills.
Expect the problem to include at least one extra force (wind, drag) beyond gravity, because pure gravity problems are too straightforward for a Calculus III exam.
Always state the acceleration vector first and work from there. Starting from velocity or position without writing a(t) is error-prone.
The final numerical answer usually requires a calculator. Make sure your symbolic expression is correct before reaching for one.
True or false: Integrating the acceleration vector once gives the position vector.
False. One integration gives velocity; you integrate again to get position.
Fill in the blank: The initial velocity of a projectile launched at angle theta with speed v_0 is <_______, _______>.
True or false: Horizontal wind changes the time of flight of a projectile launched from ground level.
False (in this model). Time of flight depends only on the vertical component.
Fill in the blank: Each component of the velocity vector gets its own _______ when integrating acceleration.
Constant of integration
True or false: To find where a projectile lands, set the x-component of s(t) to zero.
False. Set the y-component (height) to zero.
Q: A ball is thrown from ground level at 45 degrees with speed 100 ft/s. Gravity is 32 ft/s^2. Write the acceleration, velocity and position vectors (no wind).
A: a(t) = <0, -32>. v(t) = <100 cos(45), -32t + 100 sin(45)> = <50 sqrt(2), -32t + 50 sqrt(2)>. s(t) = <50 sqrt(2) t, -16t^2 + 50 sqrt(2) t>.
Q: Using the position vector above, find the time the ball lands.
A: Set y = 0: -16t^2 + 50 sqrt(2) t = 0, so t(50 sqrt(2) - 16t) = 0. Landing time t = 50 sqrt(2)/16 = 25 sqrt(2)/8 seconds.
Q: A projectile has a(t) = <-1, -10>. v(0) = <20, 15>. Find v(t).
A: v(t) = <-t + C_1, -10t + C_2>. At t = 0:
Q: Why does the vector approach handle wind more naturally than the scalar approach?
A: Wind adds a constant to the horizontal acceleration component. In the vector approach, this just changes one entry in a(t) and propagates through the integration automatically, without needing a separate parallel calculation.
Q: In the cannonball example, what is the landing time?
A: t = 747/32 seconds (approximately 23.3 seconds). This comes from setting -16t^2 + (747/2)t = 0 and taking the positive root.
This topic builds on the integration of vector-valued functions from earlier in Chapter 13 and connects to the arc length and curvature material (Section 13.3), since the velocity vector r'(t) is the same object used in arc length computations. The decomposition of acceleration into tangential and normal components (a_T and a_N) uses the TNB frame from Section 13.3, closing the loop between geometry and physics.
projectile motion vectors, vector equations of motion, cannonball problem, integration of acceleration, initial velocity decomposition, launch angle, landing time, wind resistance projectile, constant acceleration vector, MATH 163, Section 13.4, calculus 3 projectile, vector kinematics, position velocity acceleration vectors