Question

The projectile is fired at an angle of 45° to the horizontal with an initial speed of $100$per second. Find the range R and maximum height H.

Short answer

The range R is$31·05$and height H is$0·77$.

Step-by-step solution

Step 1. Given information.

Consider the given value the initial speed is $100$feet per sec and$\theta =45°$.

Step 2. Formula used.

The formula used to find range is.

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

Step 3. Finding range and height.

Substitute the values in formula where ${\nu }_o$is 100 and $\theta$is $45°$and$g=9·8$.

$$\begin{gathered} R\left(\theta \right)=\frac{{{\nu }_o}^2\sin \left(2\theta \right)}{g} \\ R\left(\theta \right)=\frac{{\left(100\right)}^2\sin \left(2\times 45°\right)}{9·8} \\ R\left(\theta \right)=31·05 \end{gathered}$$

Further simplify.

$$\begin{gathered} H\left(\theta \right)=\frac{100\times {\left(\sin \theta \right)}^2}{2g} \\ H\left(\theta \right)=\frac{100\times {\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2}{2\times 9·8} \\ H\left(\theta \right)=0·776 \end{gathered}$$