Question

The projectile is fired at an angle of 50° to the horizontal with an initial speed of 200 feet per second.Find the range and height.

Short answer

The range is 4019.5 and height is 1150041 .

Step-by-step solution

Step 1. Given information .

Consider the given initial speed 200 meter per second and angle is$50°$.

Step 2. Formula used .

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

. Finding range and height.

Substitute the values in formula where initial speed ${\nu }_o=200,\theta =50°$.

$$\begin{gathered} R\left(\theta \right)=\frac{{{\nu }_o}^2\left(\sin 2\theta \right)}{g} \\ R\left(\theta \right)=\frac{{\left(200\right)}^2\left(\sin 2\times 50°\right)}{9·8} \\ R\left(\theta \right)=4019·5 \end{gathered}$$

Further simplify.

$$\begin{gathered} H\left(\theta \right)=\frac{{{\nu }_o}^2{\left(\sin \theta \right)}^2}{2g} \\ H\left(\theta \right)=\frac{{\left(200\right)}^2{\left(\sin 50°\right)}^2}{2\times 9·8} \\ H\left(\theta \right)=\frac{40000\times {\left(0·766\right)}^2}{19·6} \\ H\left(\theta \right)=1150041 \end{gathered}$$