Question

The projectile is fired at an angle of 30° to the horizontal with an initial speed of $150$meters per second . Find the range R and height.

Short answer

The range is $1988.2$and the height is $183·75$.

Step-by-step solution

Step 1. Given information.

Consider the given value${\nu }_o=150$and $\theta$is $30°$.

Step 2. Formula used.

The formula used to find range is$R\left(\theta \right)=\frac{{{\nu }_o}^2\sin \left(2\theta \right)}{g}$and to find height isrole="math" localid="1646585858802" $H\left(\theta \right)=\frac{{{\nu }_o}^2{\left(\sin \theta \right)}^2}{2g}$

Step 3. Finding the range and height.

Substitute the values in formula where ${\nu }_o=150,\theta =30°$

$$\begin{gathered} R\left(\theta \right)=\frac{{{\nu }_o}^2\sin \left(2\theta \right)}{2g} \\ R\left(\theta \right)=\frac{22500\times {\displaystyle \frac{\sqrt{3}}{2}}}{9·8} \\ R\left(\theta \right)=1988·2 \end{gathered}$$

Further simplify.

$$\begin{gathered} H\left(\theta \right)=\frac{{{\nu }_o}^2{\left(\sin \theta \right)}^2}{{}_{2g}} \\ H\left(\theta \right)=\frac{150\times {\displaystyle \frac{1}{4}}}{2\times 9·8} \\ H\left(\theta \right)=183·7 \end{gathered}$$