Question

You are given the equations that are used to solve a problem. For each of these, you are to

a. Write a realistic problem for which these are the correct equations. Be sure that the answer your problem requests is consistent with the equations given.

b. Finish the solution of the problem, including a pictorial representation.

$$\begin{gathered} 100\mathrm{m}=0\mathrm{m}+(50\mathrm{cos\theta }){\mathrm{t}}_1 \\ 0\mathrm{m}=0\mathrm{m}+(50\mathrm{sin\theta }){\mathrm{t}}_1-\frac{1}{2}(9.8){{\mathrm{t}}_1}^2 \end{gathered}$$

Short answer

b.$$\begin{gathered} {\mathrm{t}}_1=2.041\mathrm{s} \\ \mathrm{\theta }=11.5° \end{gathered}$$

Step-by-step solution

Given information

Given equations are:

$$\begin{gathered} 100\mathrm{m}=0\mathrm{m}+(50\mathrm{cos\theta }){\mathrm{t}}_1 \\ 0\mathrm{m}=0\mathrm{m}+(50\mathrm{sin\theta }){\mathrm{t}}_1-\frac{1}{2}(9.8){{\mathrm{t}}_1}^2 \end{gathered}$$

a. We have to write a realistic problem for which these are the correct equations.

b. Finish the solution of the problem, including a pictorial representation.

Part a) Step 1: Realistic Problem

A car travels $100\mathrm{m}$with the velocity of $50\mathrm{m}/\mathrm{s}$. Calculate the time a car required to travel and also find angle of projection with which car move?

Part b) Step 1: Calculation

By using first equation:

$$\begin{gathered} 100\mathrm{m}=0\mathrm{m}+(50\mathrm{cos\theta }){\mathrm{t}}_1 \\ 100\mathrm{m}=(50\mathrm{cos\theta }){\mathrm{t}}_1 \\ {\mathrm{t}}_1=\frac{100\mathrm{m}}{50\mathrm{cos\theta }} \end{gathered}$$

Now put it in equation 2.

localid="1649071168324" $$\begin{gathered} 0\mathrm{m}=0\mathrm{m}+(50\mathrm{sin\theta }){\mathrm{t}}_1-\frac{1}{2}(9.8){{\mathrm{t}}_1}^2 \\ 0=(50\mathrm{sin\theta }){\mathrm{t}}_1-\frac{1}{2}(9.8){{\mathrm{t}}_1}^2 \\ 0=(50\mathrm{sin\theta })\frac{100}{50\mathrm{cos\theta }}-\frac{1}{2}(9.8){\left(\frac{100}{50\mathrm{cos\theta }}\right)}^2 \\ 0=\frac{100\mathrm{sin\theta }}{\mathrm{cos\theta }}-\frac{19.6}{{\cos }^2\mathrm{\theta }} \\ \frac{100\mathrm{sin\theta }}{\mathrm{cos\theta }}=\frac{19.6}{{\cos }^2\mathrm{\theta }} \\ \frac{100\mathrm{sin\theta }}{\mathrm{cos\theta }}\times {\cos }^2\mathrm{\theta }=19.6 \\ 100\mathrm{sin\theta }\times \mathrm{cos\theta }=19.6 \\ \mathrm{sin\theta }\times \mathrm{cos\theta }=0.196 \\ 2\mathrm{sin\theta }\times \mathrm{cos\theta }=2\times 0.196 \\ 2\mathrm{\theta }={\sin }^{-1}(0.392) \\ \mathrm{\theta }=\frac{23.079}{2} \\ \mathrm{\theta }=11.54° \end{gathered}$$

For time:

$$\begin{gathered} {\mathrm{t}}_1=\frac{100\mathrm{m}}{50\cos 11.54°} \\ {\mathrm{t}}_1=2.04\mathrm{s} \end{gathered}$$

Part b) Step 2: Pictorial representation 

The pictorial representation is below: