Q 123. - Precalculus Enhanced with Graphing Utilities

Question

The projectile is fired at an angle of 25° to the horizontal with an initial speed of $500$ meters per second. Find the range R and height H.

Step-by-Step Solution

Answer

The range is $39·08$ and height is $218773·6$ .

Step 1. Given information .

Consider the given initial speed that is $500$ meter per second and $\theta =25°$ .

Step 2. Formula used.

The formula used to find range R is$R\left(\theta \right)=\frac{{{\nu }_o}^2 \sin \left(2\theta \right)}{g}$ and to find height H is $H\left(\theta \right)=\frac{{{\nu }_o}^2 {\left(\sin \theta \right)}^2}{2g}$ .

Step 3. Finding the range and height.

Substitute the values ${\nu }_o=500$ and $\theta =25°$ in the formula .

$$\begin{gathered} R\left(\theta \right)=\frac{{{\nu }_o}^2 \sin \left(2\theta \right)}{g} \\ R\left(\theta \right)=\frac{{\left(500\right)}^2\times \sin 50°}{9·8} \\ R\left(\theta \right)=\frac{250000\times 0·766}{9·8} \\ R\left(\theta \right)=39·08 \end{gathered}$$

Further simplify.

$$\begin{gathered} H\left(\theta \right)=\frac{{{\nu }_o}^2{\left(\sin \theta \right)}^2}{2g} \\ H\left(\theta \right)=\frac{250000\times {\left(0·4226\right)}^2}{2\times 9·8} \\ H\left(\theta \right)=218773·6 \end{gathered}$$