Question

A football is kickedat 23.4 m/s at a38.5° angle. How faraway does it land?(Unit = m)te

Answer 1

Given:

Initial velocity = 24.3 m/s

Angle = 38.5°

Let's find how far away the ball landed.

To find the distance, apply the projectile motion formula below:

`R=(V^2\sin (2\theta))/(g)`

Where:

g = 9.8 m/s^2

V = 23.4 m/s

θ = 38.5°

Thus, we have:

`R=(23.4^2\ast\sin (2\ast38.5))/(9.8)`

Solving further:

`\begin{gathered} R=(547.56\sin(77))/(9.8) \\ \\ R=(547.56\cdot0.97437006)/(9.8) \\ \\ R=(533.52607267)/(9.8) \\ \\ R=54.4\text{ m} \end{gathered}`

Therefore, the footabll lands 54.4 meters away

ANSWER:

54.4 m