Explanation:
Let's establish our knowable variables
[tex]v_{o}=30 \frac{m}{s}[/tex], ∅=30º
[tex]x=100m[/tex]
a) Since we know the x distance that the cannonball hits the ground and the velocity at the x-axis is constant, meaning that the acceleration at x is 0.
We can calculate the time it takes the cannonball to hit the ground by using the following formula:
[tex]x=v_{x}*t[/tex]
[tex]t=\frac{x}{v_{x} }=\frac{100m}{30\frac{m}{s} cos(30)}=3.849 s[/tex]
b) To calculate the height of the castle wall, we need to find the initial height
Since the ball hits the ground at the end of it's motion, we know that y=0
[tex]y=y_{o}+v_{oy}*t+\frac{1}{2}gt^{2}[/tex]
[tex]0=y_{o}+30sin(30)*(3.849)+\frac{1}{2}(-9.81)(3.849)^{2}[/tex]
Solving for [tex]y_{o}[/tex]
[tex]y_{0}= -30sin(30)*(3.849)-\frac{1}{2}(-9.81)(3.849)^{2}=14.8575m[/tex]
c) At the highest point of its trajectory, we know that the cannonball stops ascending and began going down, that means:
[tex]v_{y}=0[/tex]
[tex]v_{x}=30cos(30)=15\sqrt{3}\frac{m}{s}[/tex]
d) Since the velocity in x is constant, the velocity just before the cannonball hits the ground is:
[tex]v_{x}=30cos(30)=15\sqrt{3}[/tex]
To find y-velocity of the cannon ball's velocity we need to use the formula:
[tex]v_{y} ^{2} =v_{oy} ^{2} +2g(y-y_{o})[/tex]
[tex]v_{y} =\sqrt{v_{oy} ^{2} +2g(y-y_{o})} =\sqrt{(30sin(30))^{2} +2(-9.8)(0-14.8575)}[/tex]
[tex]v_{y}=22.7202 \frac{m}{s}[/tex]
e) It's attached
