ToughSTEM
ToughSTEM
A question answer community on a mission
to share Solutions for all STEM major Problems.
Cant find a problem on ToughSTEM?
0
A cannon tilted up at a 26 degree angle fires a cannon ball at 71m/s from atop a 12m -high fortress wall.
What is the ball's impact speed on the ground below?

Edit
Community
1
Comment
Solutions
0
Total E = KEi + PEi
TotalE = KEf +PEf
PE = mgh
KE = 1/2mv^2

So Total E is always kept constant. it never changes. Thus it's equal to the KE + PE. Well, the PE of the cannon and thus the ball initially is mg(12). And the initial KE will be 1/2m(71)^2 **KE is a scalar quantity here**

Thus, the Total E of the ball will be 1/2m(71)^2 + m(9.8)(12)

That Total E is kept constant from start to finish (well, ignoring air resistance and whatnot). So thus at the end of its trip when the PE is equal to zero, the total E will be equal to the final KE

Total E = PEf + KEf
TotalE = 0 + 1/2mvf^2

Total E = TotalE
1/2m(71)^2 + m(9.8)(12) = 1/2mvf^2
**now you can see why the mass of the ball is neglected**
1/2(71)^2 + 117.6 = 1/2vf^2
71^2 + 2*117.6 = vf^2

vf^2 = 5276.2

vf = 72.637
Edit
Community
1
Comment
Close

Choose An Image
or
Get image from URL
GO
Close
Back
Close
What URL would you like to link?
GO
α
β
γ
δ
ϵ
ε
η
ϑ
λ
μ
π
ρ
σ
τ
φ
ψ
ω
Γ
Δ
Θ
Λ
Π
Σ
Φ
Ω
Copied to Clipboard

to interact with the community. (That's part of how we ensure only good content gets on ToughSTEM)
OR
OR
ToughSTEM is completely free, and its staying that way. Students pay way too much already.
Almost done!