Answer:
240.46 m
Explanation:
If the point in x where the package was dropped is x = 0. We can find the value of x where the package ends with the formula:
[tex]x=v_{o}t[/tex]
[tex]v_{o}[/tex] is the initial velocity, in this case is 49m/s
t is the time it takes for the package to get to the ground, wich we can found with the next formula:
[tex]y = h - \frac{1}{2}gt^2[/tex]
where h is the height: 118 m
and [tex]g = 9.8m/s^2[/tex]
solving for t:
[tex]t =\sqrt{\frac{y-h}{-\frac{1}{2} g} }[/tex]
since the ground is the point where y = 0, we have:
[tex]t=\sqrt{\frac{2h}{g} }[/tex]
[tex]t=\sqrt{\frac{2(118m)}{9.8m/s^2}} = 4.91s[/tex]
Going back to the first formula for the distance x:
[tex]x=v_{o}t = (49m/s)(4.91s)= 240.46m[/tex]
