Answer: 473.640 m
Explanation:
This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]V_{o}=48.1 m/s[/tex] is the cannonball's initial velocity
[tex]\theta=0[/tex] because we are told the cannonball is shot horizontally
[tex]t[/tex] is the time since the cannonball is shot until it hits the ground
y-component:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (2)
Where:
[tex]y_{o}=1.5m[/tex] is the initial height of the cannonball
[tex]y=0[/tex] is the final height of the cannonball (when it finally hits the ground)
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it [tex]X_{max}[/tex] and this occurs when [tex]y=0[/tex].
So, firstly we will find the time with (2):
[tex]0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2[/tex] (3)
Rearranging the equation:
[tex]0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m[/tex] (4)
[tex]-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0[/tex] (5)
This is a quadratic equation (also called equation of the second degree) of the form [tex]at^{2}+bt+c=0[/tex], which can be solved with the following formula:
[tex]t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex] (6)
Where:
[tex]a=-4.9 m/s^{2}[/tex]
[tex]b=48.1 m/s[/tex]
[tex]c=1.5 m[/tex]
Substituting the known values:
[tex]t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)}[/tex] (7)
Solving (7) we find the positive result is:
[tex]t=9.847 s[/tex] (8)
Substituting this value in (1):
[tex]x=(48.1 m/s)cos(0\°) (9.847 s)[/tex] (9)
[tex]x=473.640 m[/tex] This is the horizontal distance the cannonball traveled before it landed on the ground.
