- #1
AudioGeek
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Homework Statement
A cannon is discharged towards a castle tower (near the time when the ancient world was
meeting the modern world). The cannon ball flies 5 m above the top of the 10 metre tall tower and lands on level ground (same level as cannon) behind the tower. The cannon was positioned 300 m from the tower and was angled at 50[tex]\circ[/tex] to the horizontal. What was the muzzle velocity of the cannon ball? How far behind the tower did it land? [up + and right +]
Homework Equations
[tex]\Delta[/tex]d[tex]_{x}[/tex] = v[tex]_{1x}[/tex][tex]\Delta[/tex]t
[tex]\Delta[/tex]d[tex]_{y}[/tex] = v[tex]_{1y}[/tex][tex]\Delta[/tex]t - [tex]\frac{1}{2}[/tex]g[tex]\Delta[/tex]t[tex]^{2}[/tex]
v[tex]_{1y}[/tex] = v[tex]_{sin}[/tex][tex]\Theta[/tex]
v[tex]_{1x}[/tex] = v[tex]_{cos}[/tex][tex]\Theta[/tex]
The Attempt at a Solution
In my attempts to find the muzzle velocity I figured I could use the equation
[tex]\Delta[/tex]d[tex]_{x}[/tex] = v[tex]_{1x}[/tex][tex]\Delta[/tex]t and isolate the variable of time. Then I figured I could plug it into the equation
[tex]\Delta[/tex]d[tex]_{y}[/tex] = v[tex]_{1y}[/tex][tex]\Delta[/tex]t - [tex]\frac{1}{2}[/tex]g[tex]\Delta[/tex]t[tex]^{2}[/tex]
So I rearranged the whole thing fully and correctly to solve for [tex]\vec{v}[/tex]. After plugging in my numbers and such however, I ended up with ~35.5m/s where I should have ~56m/s. I have no idea what I'm doing wrong, please help, someone?
EDIT: Don't know why my subscripts always turn out as superscripts...