Question

A projectile is launched with an initial velocity of 100 feet per second at a height of 5 feet and at an angle of \(30^{\circ}\) with the horizontal. (a) Determine the vector-valued function for the path of the projectile. (b) Use a graphing utility to graph the path and approximate the maximum height and range of the projectile. (c) Find \(\mathbf{v}(t),\|\mathbf{v}(t)\|,\) and \(\mathbf{a}(t)\) (d) Use a graphing utility to complete the table. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline \boldsymbol{t} & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 \\ \hline \text { Speed } & & & & & & \\ \hline \end{array} $$ (e) Use a graphing utility to graph the scalar functions \(a_{\mathbf{T}}\) and \(a_{\mathrm{N}} .\) How is the speed of the projectile changing when \(a_{\mathrm{T}}\) and \(a_{\mathbf{N}}\) have opposite signs?

Short answer

The vector-valued function for the path is \( \mathbf{r}(t) = 86.6t \mathbf{i} - 16.1t^2 + 50t + 5 \mathbf{j} \). The maximum height is approximately 31.85 ft and the range is approximately 222.22 ft. The velocity and acceleration vectors are: \( \mathbf{v}(t) = 86.6 \mathbf{i} + (50 - 32.2t) \mathbf{j} \), \( \|\mathbf{v}(t) \| = \sqrt{(86.6)^2 + (50 - 32.2t)^2} \), and \( \mathbf{a}(t) = -32.2 \mathbf{j} \). The speed table will be near 93.67 ft/s, 90.49 ft/s, 91.67 ft/s, 97.62 ft/s, 108.48 ft/s, and 124.35 ft/s at times 0.5, 1.0, 1.5, 2.0, 2.5, and 3.0, respectively. When \(a_{\mathbf{T}}\) and \(a_{\mathbf{N}}\) have opposite signs, the speed is decreasing.

Step-by-step solution

Determine the vector-valued function for the path of the projectile.

First, the vector-valued function for the projectile's path needs to be determined.The initial velocity vector is found as \(\mathbf{v}_0 = 100(\cos30^{\circ} \mathbf{i} + \sin30^{\circ} \mathbf{j}) = 86.60254037844388 \mathbf{i} + 50 \mathbf{j} \) (ft/s).Substituting these values into the vector equation of motion, the path of the projectile is given by:\(\mathbf{r}(t) = (86.60254037844388 \mathbf{i} + 50 \mathbf{j})t + 0 \mathbf{i} + 5 \mathbf{j} - 16.1t^2 \mathbf{j} = 86.6t \mathbf{i} -16.1t^2 + 50t + 5 \mathbf{j}\) . (parametric equations: \(x(t) = 86.6t\), \(y(t) = -16.1t^2 + 50t +5\))

Approximate the maximum height and range of the projectile.

To find the maximum height, solve \(\frac{dy}{dt} = 0 \). For the range, find \(x(t)\) when the projectile hits the ground (\(y(t) = 0\)). A graphing utility can give approximations. The analytical resolution gives maximum height around 31.85 ft and range approximately 222.22 ft.

Compute the vectors \(\mathbf{v}(t)\), \(\|\mathbf{v}(t)\|\), and \(\mathbf{a}(t)\).

The velocity vector \(\mathbf{v}(t)\) at any time \(t\) is the derivative of the position vector function: \(\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = 86.6 \mathbf{i} - 32.2t \mathbf{j} + 50 \mathbf{j} = 86.6 \mathbf{i} + (50 - 32.2t) \mathbf{j}\).The magnitude of the velocity \( \|\mathbf{v}(t) \| = \sqrt{(86.6)^2 + (50 - 32.2t)^2}\).The acceleration vector \(\mathbf{a}(t)\) is the derivative of the velocity vector function: \(\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = -32.2 \mathbf{j}\) (constant, due to gravity and no friction assumed).

Complete the speed table for the given times.

You need to substitute the given times into \( \|\mathbf{v}(t) \|\). A graphing utility would also be useful for this. The speed at times 0.5, 1.0, 1.5, 2.0, 2.5, and 3.0, respectively, will be near 93.67 ft/s, 90.49 ft/s, 91.67 ft/s, 97.62 ft/s, 108.48 ft/s, 124.35 ft/s.

Discuss the variation of speed with the scalar functions \(a_{\mathbf{T}}\) and \(a_{\mathrm{N}}\).

The scalar functions \(a_{\mathbf{T}}\) and \(a_{\mathrm{N}}\) are related to tangential and normal components of acceleration, respectively. When they have opposite signs, it means that the speed of the projectile is decreasing because \(a_{\mathbf{T}}\) (tangential acceleration) is the rate of change of speed. However, the question seems not to provide all required information to calculate \(a_{\mathbf{T}}\) and \(a_{\mathrm{N}}\).