Question

A football player punts a ball at an angle of 47° at 1.5 feet off the ground, with an initial velocity of 74 feet per second. How far away does the ball land, in feet, when hitting the ground?

Answer 1

Explanation:

To solve the question, we will have to first of all list out the given parameters

$\begin{gathered} u=74ft\text{/sec} \\ \theta=47^0 \\ S=1.5ft \end{gathered}$

The question asked that we obtain how far away does the ball land, in feet, when hitting the ground

This means we have to obtain the range

The formula to get the range is given by

$\begin{gathered} \text{Range}=\frac{u^2\sin 2\theta}{g} \\ \text{where} \\ g=32.17ft\text{/sec}^2 \end{gathered}$

Thus

$\begin{gathered} \text{Range}=\frac{74^2\sin (2\times47)}{32.17} \\ \\ \text{Range}=169.81ft \end{gathered}$

Hence, we can conclude that the ball will be about 168.376 feet away