Chapter MPS, Problem 14.3MPS

To determine

Verify the identity $\frac{{\upsilon }^2{\sin }^2\theta }{2g}=\frac{{\upsilon }^2{\cos }^2\theta }{2g{\cot }^2\theta }$ .

Answer to Problem 14.3MPS

The identity is verified.

Explanation of Solution

Given information:

In the formula $h=\frac{{\upsilon }^2{\sin }^2\theta }{2g},h$ is the maximum height reached by a rocket, $\theta$ is the angle between the ground and the initial path of the object, $\upsilon$ is the rocket’s initial velocity, and $g$ is the acceleration due to gravity.

Calculation:

Condider the equation:

  $h=\frac{v^2{\sin }^2\theta }{2g}$

Verify the identity as follows:

  $\begin{array}{l}\frac{v^2{\sin }^2\theta }{2g}\stackrel{?}{{\displaystyle =}}\frac{v^2{\cos }^2\theta }{2g{\cot }^2\theta }\\ \frac{v^2{\sin }^2\theta }{2g}\stackrel{?}{{\displaystyle =}}\frac{v^2{\cos }^2\theta }{2g.{\left(\frac{\cos \theta }{\sin \theta }\right)}^2}\\ \frac{v^2{\sin }^2\theta }{2g}\stackrel{?}{{\displaystyle =}}\frac{v^2{\cos }^2\theta }{2g\cdot \frac{{\cos }^2\theta }{{\sin }^2\theta }}\\ \frac{v^2{\sin }^2\theta }{2g}\stackrel{?}{{\displaystyle =}}\frac{v^2{\cos }^2\theta }{2g}\times \frac{{\sin }^2\theta }{{\cos }^2\theta }\\ \frac{v^2{\sin }^2\theta }{2g}=\frac{v^2{\sin }^2\theta }{2g}\end{array}$

Hence, the identity is verified.