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4. Potato launch. A potato is launched from a potato cannon held 1 meter above ...
Sep 26, 2024
Generated Graph
Solution by Steps
step 1
To find the velocity vector function v(t) \mathbf{v}(t) , we integrate the acceleration vector a(t)=⟨0,−mg⟩ \mathbf{a}(t) = \langle 0, -mg \rangle :
\mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \langle v_x(0), v_y(0) - mg t \rangle
step 2
Given the initial velocity v(0)=⟨90cos⁡(30∘),90sin⁡(30∘)⟩ \mathbf{v}(0) = \langle 90 \cos(30^{\circ}), 90 \sin(30^{\circ}) \rangle , we find:
v_x(0) = 90 \cos(30^{\circ}) = 90 \cdot \frac{\sqrt{3}}{2} = 45\sqrt{3} \quad \text{and} \quad v_y(0) = 90 \sin(30^{\circ}) = 90 \cdot \frac{1}{2} = 45
step 3
Thus, the velocity vector function becomes:
\mathbf{v}(t) = \langle 45\sqrt{3}, 45 - mgt \rangle
step 4
To find the position vector function r(t) \mathbf{r}(t) , we integrate the velocity vector function:
\mathbf{r}(t) = \int \mathbf{v}(t) \, dt = \langle 45\sqrt{3}t, 45t - \frac{1}{2}mgt^2 \rangle + \mathbf{r}(0)
step 5
Assuming the initial position r(0)=⟨0,1⟩ \mathbf{r}(0) = \langle 0, 1 \rangle , we have:
\mathbf{r}(t) = \langle 45\sqrt{3}t, 45t - \frac{1}{2}mgt^2 + 1 \rangle
Answer
The velocity vector function is v(t)=⟨453,45−mgt⟩ \mathbf{v}(t) = \langle 45\sqrt{3}, 45 - mgt \rangle and the position vector function is r(t)=⟨453t,45t−12mgt2+1⟩ \mathbf{r}(t) = \langle 45\sqrt{3}t, 45t - \frac{1}{2}mgt^2 + 1 \rangle
Key Concept
The motion of a projectile under constant gravitational acceleration can be described using vector functions for velocity and position.
Explanation
The derived functions represent the velocity and position of the potato as it moves through the gravitational field, illustrating the effects of initial velocity and gravitational acceleration.
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