Motion in Space and Curvature
Velocity and Acceleration
If the position of an object in space is described by a vector-valued function r ⃗ ( t ) = ⟨ x ( t ) , y ( t ) , z ( t ) ⟩ r⃗ (t)=⟨x(t),y(t),z(t)⟩ r ⃗ ( t ) = ⟨ x ( t ) , y ( t ) , z ( t ) ⟩
Velocity: v ⃗ ( t ) = r ⃗ ′ ( t ) Speed: ∣ v ⃗ ( t ) ∣ = x ′ ( t ) 2 + y ′ ( t ) 2 + z ′ ( t ) 2 Acceleration: a ⃗ ( t ) = v ⃗ ′ ( t ) = r ⃗ ′ ′ ( t ) \begin{aligned}
\textrm{Velocity: } &\vec{v}(t) = \vec{r}\ '(t)\
\textrm{Speed: } &|\vec{v}(t)| = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}\
\textrm{Acceleration: } &\vec{a}(t) = \vec{v}\ '(t) = \vec{r}\ ''(t)
\end{aligned} Velocity: Speed: Acceleration: v ( t ) = r ′ ( t ) ∣ v ( t ) ∣ = x ′ ( t ) 2 + y ′ ( t ) 2 + z ′ ( t ) 2 a ( t ) = v ′ ( t ) = r ′ ′ ( t )
Note that the velocity is a vector, and the speed is a scalar (the magnitude of the velocity vector).
If r ⃗ ( t ) = t i ^ + t 3 j ^ + t 2 k ^ \vec{r}(t)=t\ \hat{i} + t^3\ \hat{j} + t^2\ \hat{k} r ( t ) = t i ^ + t 3 j ^ + t 2 k ^ v ⃗ ( t ) \vec{v}(t) v ( t ) a ⃗ ( t ) \vec{a}(t) a ( t )
Solution
Find the first and second derivatives of r ⃗ ( t ) r⃗ (t) r ⃗ ( t ) v ⃗ ( t ) = i ^ + 3 t 2 j ^ + 2 t k ^ a ⃗ ( t ) = 6 t j ^ + 2 k ^ \ans{\begin{aligned}
\vec{v}(t) &= \hat{i} + 3t^2\ \hat{j} + 2t\ \hat{k}\
\vec{a}(t) &= 6t\ \hat{j} + 2\ \hat{k}
\end{aligned}} v ( t ) a ( t ) = i ^ + 3 t 2 j ^ + 2 t k ^ = 6 t j ^ + 2 k ^
If r ⃗ ( t ) = 3 cos t i ^ + 3 sin t j ^ \vec{r}(t)=3\cos t\ \hat{i} + 3\sin t\ \hat{j} r ( t ) = 3 cos t i ^ + 3 sin t j ^ v ⃗ ( t ) v⃗ (t) v ⃗ ( t ) ∣ v ⃗ ( t ) ∣ |v⃗ (t)| ∣ v ⃗ ( t ) ∣ a ⃗ ( t ) a⃗ (t) a ⃗ ( t )
Solution
Note that x ( t ) 2 + y ( t ) 2 = 9 x(t)^2 + y(t)^2 = 9 x ( t ) 2 + y ( t ) 2 = 9 x y xy x y 3 3 3
v ⃗ ( t ) = − 3 sin t i ^ + 3 cos t j ^ ⟶ ∣ v ⃗ ( t ) ∣ = 3 a ⃗ ( t ) = − 3 cos t i ^ − 3 sin t j ^ \ans{\begin{aligned}
\vec{v}(t) &= -3\sin t\ \hat{i} + 3\cos t\ \hat{j} \longrightarrow |\vec{v}(t)| = 3\
\vec{a}(t) &= -3\cos t\ \hat{i} - 3\sin t\ \hat{j}
\end{aligned}} v ( t ) a ( t ) = − 3 sin t i ^ + 3 cos t j ^ ⟶ ∣ v ( t ) ∣ = 3 = − 3 cos t i ^ − 3 sin t j ^
Curvature
Suppose an object is traveling along a curving path through space (think of a car driving along a mountain path). We can use the unit tangent and normal vectors at each point to describe this motion.
Note that T ⃗ ⊥ N ⃗ ⟶ T ⃗ ⋅ N ⃗ = 0 \vec{T} \perp \vec{N} \longrightarrow \vec{T} \cdot \vec{N} = 0 T ⊥ N ⟶ T ⋅ N = 0
Unit Tangent Vector : we've already seen this
T ⃗ ( t ) = r ⃗ ′ ( t ) ∣ r ⃗ ′ ( t ) ∣ \ans{\vec{T}(t) = \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|}} T ( t ) = ∣ r ′ ( t ) ∣ r ′ ( t )
Curvature : how quickly the direction (tangent vector) changes
κ = ∣ T ⃗ ′ ( t ) ∣ ∣ r ⃗ ′ ( t ) ∣ = ∣ v ⃗ × a ⃗ ∣ ∣ v ⃗ ∣ 3 \ans{\kappa = \dfrac{|\vec{T}\ '(t)|}{|\vec{r}\ '(t)|} = \dfrac{|\vec{v} \times \vec{a}|}{|\vec{v}|^3}} κ = ∣ r ′ ( t ) ∣ ∣ T ′ ( t ) ∣ = ∣ v ∣ 3 ∣ v × a ∣
Unit Normal Vector : as long as κ = 0 \kappa\ \cancel=\ 0 κ = 0 N ⃗ ( t ) = T ⃗ ′ ( t ) ∣ T ⃗ ′ ( t ) ∣ \ans{\vec{N}(t) = \dfrac{\vec{T}\ '(t)}{|\vec{T}\ '(t)|}} N ( t ) = ∣ T ′ ( t ) ∣ T ′ ( t )
Curvature of a Circle
If r ⃗ ( t ) = ⟨ R cos t , R sin t ⟩ \vec{r}(t)=\langle R\cos t, R\sin t \rangle r ( t ) = ⟨ R cos t , R sin t ⟩ R R R
Solution
T ⃗ ( t ) = r ⃗ ′ ( t ) ∣ r ⃗ ′ ( t ) ∣ = 1 R ⟨ − R sin t , R cos t ⟩ = ⟨ − sin t , cos t ⟩ κ = ∣ T ⃗ ′ ( t ) ∣ ∣ r ⃗ ′ ( t ) ∣ = 1 R (note: constant curvature) \begin{aligned}
\vec{T}(t) &= \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|} = \dfrac{1}{R} \langle -R\sin t, R\cos t \rangle\
&= \ans{\langle -\sin t,\cos t \rangle}\
\
\kappa &= \dfrac{|\vec{T}\ '(t)|}{|\vec{r}\ '(t)|} = \ans{\dfrac{1}{R}} \textrm{ (note: constant curvature)}
\end{aligned} T ( t ) κ = ∣ r ′ ( t ) ∣ r ′ ( t ) = R 1 ⟨ − R sin t , R cos t ⟩ = ⟨ − sin t , cos t ⟩ = ∣ r ′ ( t ) ∣ ∣ T ′ ( t ) ∣ = R 1 (note: constant curvature)
Helix
A specific helix is defined by r ⃗ ( t ) = ⟨ 2 cos t , 2 sin t , 3 t ⟩ \vec{r}(t)=\langle 2\cos t, 2\sin t, 3t \rangle r ( t ) = ⟨ 2 cos t , 2 sin t , 3 t ⟩
Find the curvature, unit tangent, and unit normal.
Solution
Curvature: κ = ∣ v ⃗ × a ⃗ ∣ ∣ v ⃗ ∣ 3 \kappa = \dfrac{|\vec{v} \times \vec{a}|}{|\vec{v}|^3} κ = ∣ v ∣ 3 ∣ v × a ∣
v ⃗ = ⟨ − 2 sin t , 2 cos t , 3 ⟩ a ⃗ = ⟨ − 2 cos t , − 2 sin t , 0 ⟩ v ⃗ × a ⃗ = ∣ i ^ j ^ k ^ − 2 sin t 2 cos t 3 − 2 cos t − 2 sin t 0 ∣ = 6 sin t i ^ − 6 cos t j ^ + 4 k ^ ∣ v ⃗ × a ⃗ ∣ = 36 sin 2 t + 36 cos 2 t + 16 = 36 + 16 = 2 13 ∣ v ⃗ ∣ 3 = 4 sin 2 t + 4 cos 2 t + 9 = 4 + 9 = 13 ⟹ κ = ∣ v ⃗ × a ⃗ ∣ ∣ v ⃗ ∣ 3 = 2 13 13 13 = 2 13 \begin{aligned}
\vec{v} &= \langle -2\sin t,2\cos t,3 \rangle\
\vec{a} &= \langle -2\cos t,-2\sin t,0 \rangle\
\
\vec{v} \times \vec{a} &= \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\
-2\sin t & 2\cos t & 3\
-2\cos t & -2\sin t & 0
\end{vmatrix}\
\
&= 6\sin t\ \hat{i} - 6\cos t\ \hat{j} + 4\ \hat{k}\
\
|\vec{v} \times \vec{a}| &= \sqrt{36\sin^2 t + 36\cos^2 t + 16} = \sqrt{36+16} = 2\sqrt{13}\
\
|\vec{v}|^3 &= \sqrt{4\sin^2 t + 4\cos^2 t + 9} = \sqrt{4+9} = \sqrt{13}\
\
&\implies \ans{\kappa = \dfrac{|\vec{v} \times \vec{a}|}{|\vec{v}|^3} = \dfrac{2\sqrt{13}}{13\sqrt{13}} = \dfrac{2}{13}}
\end{aligned} v a v × a ∣ v × a ∣ ∣ v ∣ 3 = ⟨ − 2 sin t , 2 cos t , 3 ⟩ = ⟨ − 2 cos t , − 2 sin t , 0 ⟩ = ∣ ∣ ∣ ∣ ∣ ∣ i ^ − 2 sin t − 2 cos t j ^ 2 cos t − 2 sin t k ^ 3 0 ∣ ∣ ∣ ∣ ∣ ∣ = 6 sin t i ^ − 6 cos t j ^ + 4 k ^ = 3 6 sin 2 t + 3 6 cos 2 t + 1 6 = 3 6 + 1 6 = 2 1 3 = 4 sin 2 t + 4 cos 2 t + 9 = 4 + 9 = 1 3 ⟹ κ = ∣ v ∣ 3 ∣ v × a ∣ = 1 3 1 3 2 1 3 = 1 3 2
Unit Tangent: T ⃗ ( t ) = r ⃗ ′ ( t ) ∣ r ⃗ ′ ( t ) ∣ \vec{T}(t) = \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|} T ( t ) = ∣ r ′ ( t ) ∣ r ′ ( t )
r ⃗ ′ ( t ) = ⟨ − 2 sin t , 2 cos t , 3 ⟩ ∣ r ⃗ ′ ( t ) ∣ = 13 ⟹ T ⃗ ( t ) = ⟨ − 2 13 13 sin t , 2 13 13 cos t , 3 13 13 ⟩ \begin{aligned}
\vec{r}\ '(t) &= \langle -2\sin t,2\cos t,3 \rangle\
|\vec{r}\ '(t)| &= \sqrt{13}\
\
&\implies \ans{\vec{T}(t) = \left\langle -\dfrac{2\sqrt{13}}{13}\sin t, \dfrac{2\sqrt{13}}{13}\cos t,\dfrac{3\sqrt{13}}{13} \right\rangle}
\end{aligned} r ′ ( t ) ∣ r ′ ( t ) ∣ = ⟨ − 2 sin t , 2 cos t , 3 ⟩ = 1 3 ⟹ T ( t ) = ⟨ − 1 3 2 1 3 sin t , 1 3 2 1 3 cos t , 1 3 3 1 3 ⟩
Unit Normal: N ⃗ ( t ) = T ⃗ ′ ( t ) ∣ T ⃗ ′ ( t ) ∣ \vec{N}(t) = \dfrac{\vec{T}\ '(t)}{|\vec{T}\ '(t)|} N ( t ) = ∣ T ′ ( t ) ∣ T ′ ( t ) T ⃗ ( t ) = ⟨ − 2 13 13 sin t , 2 13 13 cos t , 3 13 13 ⟩ T ⃗ ′ ( t ) = ⟨ − 2 13 13 cos t , − 2 13 13 sin t , 0 ⟩ ∣ T ⃗ ′ ( t ) ∣ = 2 13 13 ⟹ N ⃗ ( t ) = ⟨ − cos t , − sin t , 0 ⟩ \begin{aligned}
\vec{T}(t) &= \left\langle -\dfrac{2\sqrt{13}}{13}\sin t, \dfrac{2\sqrt{13}}{13}\cos t,\dfrac{3\sqrt{13}}{13} \right\rangle\
\vec{T}\ '(t) &= \left\langle -\dfrac{2\sqrt{13}}{13}\cos t, -\dfrac{2\sqrt{13}}{13}\sin t,0 \right\rangle\
|\vec{T}\ '(t)| &= \dfrac{2\sqrt{13}}{13}\
\
&\implies \ans{\vec{N}(t) = \left\langle -\cos t, -\sin t,0 \right\rangle}
\end{aligned} T ( t ) T ′ ( t ) ∣ T ′ ( t ) ∣ = ⟨ − 1 3 2 1 3 sin t , 1 3 2 1 3 cos t , 1 3 3 1 3 ⟩ = ⟨ − 1 3 2 1 3 cos t , − 1 3 2 1 3 sin t , 0 ⟩ = 1 3 2 1 3 ⟹ N ( t ) = ⟨ − cos t , − sin t , 0 ⟩
Components of Acceleration
Consider a car driving along a road; the acceleration of the car includes changes in speed (speeding up or slowing down) and changes in direction (turning), so we can talk about the tangential component of acceleration (speed changes) and the normal component of acceleration (direction changes).
a ⃗ = a N N ⃗ + a T T ⃗ where a N = a ⃗ ⋅ N ⃗ = ∣ a ⃗ × v ⃗ ∣ ∣ v ⃗ ∣ and a T = a ⃗ ⋅ T ⃗ = a ⃗ ⋅ v ⃗ ∣ v ⃗ ∣ \begin{aligned}
\vec{a} &= a_N \vec{N} + a_T \vec{T}\
\textrm{where } a_N &= \vec{a} \cdot \vec{N} = \dfrac{|\vec{a} \times \vec{v}|}{|\vec{v}|}\
\textrm{and } a_T &= \vec{a} \cdot \vec{T} = \dfrac{\vec{a} \cdot \vec{v}}{|\vec{v}|}
\end{aligned} a where a N and a T = a N N + a T T = a ⋅ N = ∣ v ∣ ∣ a × v ∣ = a ⋅ T = ∣ v ∣ a ⋅ v
Car Traveling
If r ⃗ ( t ) = ⟨ t , t 2 ⟩ \vec{r}(t)=\langle t,t^2 \rangle r ( t ) = ⟨ t , t 2 ⟩ − 2 ≤ t ≤ 2 -2 \leq t \leq 2 − 2 ≤ t ≤ 2
Solution
First, find the acceleration vector:
r ⃗ ( t ) = ⟨ t , t 2 ⟩ ⟶ v ⃗ ( t ) = ⟨ 1 , 2 t ⟩ ⟶ a ⃗ ( t ) = ⟨ 0 , 2 ⟩ \vec{r}(t) = \langle t,t^2 \rangle \longrightarrow \vec{v}(t) = \langle 1,2t \rangle \longrightarrow \vec{a}(t) = \langle 0,2 \rangle r ( t ) = ⟨ t , t 2 ⟩ ⟶ v ( t ) = ⟨ 1 , 2 t ⟩ ⟶ a ( t ) = ⟨ 0 , 2 ⟩
Next, find the unit tangent and unit normal vectors:
T ⃗ ( t ) = r ⃗ ′ ( t ) ∣ r ⃗ ′ ( t ) ∣ = ⟨ 1 , 2 t ⟩ 1 + 4 t 2 T ⃗ ′ ( t ) = ⟨ − 4 t ( 1 + 4 t 2 ) 3 / 2 , 2 1 + 4 t 2 − 8 t 2 ( 1 + 4 t 2 ) 3 / 2 ⟩ = 1 ( 1 + 4 t 2 ) 3 / 2 ⟨ − 4 t , 2 ⟩ N ⃗ ( t ) = T ⃗ ′ ( t ) ∣ T ⃗ ′ ( t ) ∣ = 1 4 t 2 + 1 ⟨ − 2 t , 1 ⟩ a N = a ⃗ ⋅ N ⃗ = 2 1 + 4 t 2 a T = a ⃗ ⋅ T ⃗ = 4 t 1 + 4 t 2 \begin{aligned}
\vec{T}(t) &= \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|} = \dfrac{\langle 1,2t \rangle}{\sqrt{1+4t^2}}\
\
\vec{T}\ '(t) &= \left\langle \dfrac{-4t}{(1+4t^2)^{3/2}}, \dfrac{2}{\sqrt{1+4t^2}}-\dfrac{8t^2}{(1+4t^2)^{3/2}} \right\rangle\
&= \dfrac{1}{(1+4t^2)^{3/2}} \langle -4t,2 \rangle\
\
\vec{N}(t) &= \dfrac{\vec{T}\ '(t)}{|\vec{T}\ '(t)|} = \dfrac{1}{\sqrt{4t^2+1}} \langle -2t,1 \rangle\
\
\end{aligned}\
\ans{\begin{aligned}
a_N &= \vec{a} \cdot \vec{N} = \dfrac{2}{\sqrt{1+4t^2}}\
a_T &= \vec{a} \cdot \vec{T} = \dfrac{4t}{\sqrt{1+4t^2}}
\end{aligned}} T ( t ) T ′ ( t ) N ( t ) = ∣ r ′ ( t ) ∣ r ′ ( t ) = 1 + 4 t 2 ⟨ 1 , 2 t ⟩ = ⟨ ( 1 + 4 t 2 ) 3 / 2 − 4 t , 1 + 4 t 2 2 − ( 1 + 4 t 2 ) 3 / 2 8 t 2 ⟩ = ( 1 + 4 t 2 ) 3 / 2 1 ⟨ − 4 t , 2 ⟩ = ∣ T ′ ( t ) ∣ T ′ ( t ) = 4 t 2 + 1 1 ⟨ − 2 t , 1 ⟩ a N a T = a ⋅ N = 1 + 4 t 2 2 = a ⋅ T = 1 + 4 t 2 4 t
(note: we could have also calculated a N a_N a N a T a_T a T
Projectile Motion
In Two Dimensions: no wind
Suppose we launch a projectile and neglect air resistance, so the only force on the projectile is gravity. Suppose the components of the initial velocity vector are
v 0 x = ∣ v ⃗ 0 ∣ cos θ v 0 y = ∣ v ⃗ 0 ∣ sin θ \begin{aligned}
v_{0x} &= |\vec{v}0| \cos \theta\
v {0y} &= |\vec{v}_0| \sin \theta
\end{aligned} v 0 x v 0 y = ∣ v 0 ∣ cos θ = ∣ v 0 ∣ sin θ
Since a ⃗ = r ⃗ ′ ′ \vec{a} = \vec{r}\ '' a = r ′ ′ r ⃗ r⃗ r ⃗
a ⃗ = − g j ^ v ⃗ = ∫ a ⃗ d t = − g t j ^ + C ⃗ ⟶ C ⃗ = v ⃗ 0 v ⃗ = − g t j ^ + v ⃗ 0 r ⃗ = ∫ v ⃗ d t = − 1 2 g t 2 j ^ + v ⃗ 0 t + C ⃗ ⟶ C ⃗ = r ⃗ 0 r ⃗ = − 1 2 g t 2 j ^ + v ⃗ 0 t + r ⃗ 0 \begin{aligned}
\vec{a} &= -g\ \hat{j}\
\
\vec{v} &= \int \vec{a} \ dt = -gt\ \hat{j} + \vec{C}\
&\longrightarrow \vec{C} = \vec{v}_0\
\vec{v} &= -gt\ \hat{j} + \vec{v}_0\
\
\vec{r} &= \int \vec{v}\ dt = -\dfrac{1}{2}gt^2\ \hat{j} + \vec{v}_0 t + \vec{C}\
&\longrightarrow \vec{C} = \vec{r}_0\
\
\vec{r} &= -\dfrac{1}{2}gt^2\ \hat{j} + \vec{v}_0 t + \vec{r}_0
\end{aligned} a v v r r = − g j ^ = ∫ a d t = − g t j ^ + C ⟶ C = v 0 = − g t j ^ + v 0 = ∫ v d t = − 2 1 g t 2 j ^ + v 0 t + C ⟶ C = r 0 = − 2 1 g t 2 j ^ + v 0 t + r 0
Summary
a ⃗ = ⟨ 0 , − g ⟩ v ⃗ = ⟨ v 0 x , − g t + v 0 y ⟩ r ⃗ = ⟨ v 0 x t + r 0 x , − 1 2 g t 2 + v 0 y t + r 0 y ⟩ \begin{aligned}
\vec{a} &= \langle 0,-g \rangle\
\
\vec{v} &= \langle v_{0x},-gt+v_{0y} \rangle\
\
\vec{r} &= \langle v_{0x}t + r_{0x}, -\dfrac{1}{2}gt^2+v_{0y}t+r_{0y} \rangle
\end{aligned} a v r = ⟨ 0 , − g ⟩ = ⟨ v 0 x , − g t + v 0 y ⟩ = ⟨ v 0 x t + r 0 x , − 2 1 g t 2 + v 0 y t + r 0 y ⟩
(note: the gravitational constant is g = 9.81 m / s 2 = 32.2 f t / s 2 g=9.81\ m/s^2 = 32.2 \ ft/s^2 g = 9 . 8 1 m / s 2 = 3 2 . 2 f t / s 2
Projectile Motion
A projectile is launched at 80 80 8 0 30 30 3 0
How long is the projectile in the air?
What is the range?
What is the maximum height?
Solution
v 0 x = 80 cos 3 0 ∘ = 40 3 v 0 y = 80 sin 3 0 ∘ = 40 r 0 x = r 0 y = 0 (by assumption) a ⃗ = ⟨ 0 , − 32 ⟩ v ⃗ = ⟨ v 0 x , − 32 t + v 0 y ⟩ = ⟨ 40 3 , − 32 t + 40 ⟩ r ⃗ = ⟨ 40 3 t , − 16 t 2 + 40 t ⟩ \begin{aligned}
v_{0x} &= 80 \cos 30^{\circ} = 40\sqrt{3}\
v_{0y} &= 80 \sin 30^{\circ} = 40\
\
r_{0x} &= r_{0y} = 0 \ \ \textrm{(by assumption)}\
\
\vec{a} &= \langle 0,-32 \rangle\
\
\vec{v} &= \langle v_{0x},-32t+v_{0y} \rangle\
&= \langle 40\sqrt{3},-32t+40 \rangle\
\
\vec{r} &= \langle 40\sqrt{3}t, -16t^2+40t \rangle
\end{aligned} v 0 x v 0 y r 0 x a v r = 8 0 cos 3 0 ∘ = 4 0 3 = 8 0 sin 3 0 ∘ = 4 0 = r 0 y = 0 (by assumption) = ⟨ 0 , − 3 2 ⟩ = ⟨ v 0 x , − 3 2 t + v 0 y ⟩ = ⟨ 4 0 3 , − 3 2 t + 4 0 ⟩ = ⟨ 4 0 3 t , − 1 6 t 2 + 4 0 t ⟩
Time in the air: find t t t j ^ \hat{j} j ^ 0 0 0 − 16 t 2 + 40 t = 0 ⟶ t ( 40 − 16 t ) = 0 ⟶ t = 0 , 40 16 ⟹ t = 5 2 s \begin{aligned}
-16t^2+40t &= 0 \longrightarrow t(40-16t) = 0 \longrightarrow t = 0,\dfrac{40}{16}\
&\implies \ans{t=\dfrac{5}{2}\ s}
\end{aligned} − 1 6 t 2 + 4 0 t = 0 ⟶ t ( 4 0 − 1 6 t ) = 0 ⟶ t = 0 , 1 6 4 0 ⟹ t = 2 5 s
Range: find the i ^ \hat{i} i ^ t = 5 2 t=\dfrac{5}{2} t = 2 5 40 3 ( 5 2 ) = 100 3 ⟹ Range ≈ 173.21 f t \begin{aligned}
40\sqrt{3}\left(\dfrac{5}{2}\right) &= 100\sqrt{3}\
&\implies \ans{\textrm{Range } \approx 173.21 \ ft}
\end{aligned} 4 0 3 ( 2 5 ) = 1 0 0 3 ⟹ Range ≈ 1 7 3 . 2 1 f t
Max Height: find where the j ^ \hat{j} j ^ 0 0 0 − 32 t + 40 = 0 ⟶ t = 40 32 = 5 4 at t = 5 4 , r ⃗ = − 16 ( 5 4 ) 2 + 40 ( 5 4 ) ⟹ Max Height = 25 f t \begin{aligned}
-32t+40 &= 0 \longrightarrow t = \dfrac{40}{32} = \dfrac{5}{4}\
\textrm{at } t &= \dfrac{5}{4}, \ \vec{r} = -16\left(\dfrac{5}{4}\right)^2+40\left(\dfrac{5}{4}\right)
&\implies \ans{\textrm{Max Height } = 25 \ ft}
\end{aligned} − 3 2 t + 4 0 at t = 0 ⟶ t = 3 2 4 0 = 4 5 = 4 5 , r = − 1 6 ( 4 5 ) 2 + 4 0 ( 4 5 ) ⟹ Max Height = 2 5 f t
In Three Dimensions: crosswind
Suppose we launch a projectile due east at 300 300 3 0 0 45 45 4 5 0.36 0.36 0 . 3 6 2 ^2 2
In three dimensions, the acceleration vector is
a ⃗ = ⟨ 0 , 0.36 , − 9.81 ⟩ \vec{a} = \langle 0,0.36,-9.81 \rangle a = ⟨ 0 , 0 . 3 6 , − 9 . 8 1 ⟩
Thus, the velocity vector is
v ⃗ = ⟨ v 0 x , 0.36 t + v 0 y , − 9.81 t + v o z ⟩ = ⟨ 150 2 , 0.36 t , − 9.81 t + 150 2 ⟩ \begin{aligned}
\vec{v} &= \langle v_{0x},0.36t+v_{0y},-9.81t+v_{oz} \rangle\
&= \langle 150\sqrt{2}, 0.36t,-9.81t+150\sqrt{2} \rangle
\end{aligned} v = ⟨ v 0 x , 0 . 3 6 t + v 0 y , − 9 . 8 1 t + v o z ⟩ = ⟨ 1 5 0 2 , 0 . 3 6 t , − 9 . 8 1 t + 1 5 0 2 ⟩
Finally, the position vector is
r ⃗ = ⟨ 150 2 t , 0.18 t 2 , − 4.9 t 2 + 150 2 t ⟩ \vec{r} = \langle 150\sqrt{2}t,0.18t^2,-4.9t^2+150\sqrt{2}t \rangle r = ⟨ 1 5 0 2 t , 0 . 1 8 t 2 , − 4 . 9 t 2 + 1 5 0 2 t ⟩
The projectile lands when the k ^ \hat{k} k ^ r ⃗ r⃗ r ⃗ 0 0 0 − 4.9 t 2 + 150 2 t = 0 ⟶ t ( 150 2 − 4.9 t ) = 0 t = 0 , 43.3 r ⃗ ( 43.3 ) = ⟨ 9183.7 , 337.4 ⟩ \begin{aligned}
-4.9t^2+150\sqrt{2}t &= 0 \longrightarrow t(150\sqrt{2}-4.9t) = 0\
t &= 0, 43.3\
\
\vec{r}(43.3) &= \ans{\langle 9183.7,337.4 \rangle}
\end{aligned} − 4 . 9 t 2 + 1 5 0 2 t t r ( 4 3 . 3 ) = 0 ⟶ t ( 1 5 0 2 − 4 . 9 t ) = 0 = 0 , 4 3 . 3 = ⟨ 9 1 8 3 . 7 , 3 3 7 . 4 ⟩
The projectile lands 9184 9184 9 1 8 4 337.4 337.4 3 3 7 . 4
