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:
[tex]R=\frac{V^2\sin (2\theta)}{g}[/tex]
Where:
g = 9.8 m/s^2
V = 23.4 m/s
θ = 38.5°
Thus, we have:
[tex]R=\frac{23.4^2\ast\sin (2\ast38.5)}{9.8}[/tex]
Solving further:
[tex]\begin{gathered} R=\frac{547.56\sin(77)}{9.8} \\ \\ R=\frac{547.56\cdot0.97437006}{9.8} \\ \\ R=\frac{533.52607267}{9.8} \\ \\ R=54.4\text{ m} \end{gathered}[/tex]
Therefore, the footabll lands 54.4 meters away
ANSWER:
54.4 m
