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July 23, 2014

Posted by **dave** on Sunday, November 29, 2009 at 3:31pm.

- college physics -
**MathMate**, Sunday, November 29, 2009 at 3:53pmFirst check if it is feasible!

The fortress is 40 m high and 100 m away.

The angle θ=tan^{-1}(40,100)=21.8°.

Since θ < angle of elevation of cannon of 37°, it is feasible.

Let the muzzle velocity be v0, and the angle of elevation be φ.

Horizontal component of velocity=v0 cos(φ)

Vertical component of velocity = v0 sin(φ)

Let Sh(t)=horizontal distance at time t after firing, and

Sv(t)=vertical distance at time t.

g=acceleration due to gravity = 9.8 m/s²

Then

Sh(t)=v0 cos(φ) t

To hit target, we get

Sh(t)=100 m = v0 cos(φ) t

t=100/(v0cos(φ)) ....... (1)

Solve for t in terms of v0 and φ.

Sv(t) = 40 = v0 sin(φ) t - (1/2)gt²

Substitute t into above equation to and solve for v0.

There should be two solutions. Any negative value of v0 should be rejected.

Each retained solution must be substituted into equation (1). Any solution which results in a negative value of t should also be rejected.

I get approximately v0=47 m/s and t between 2.5 to 3 seconds.

- college physics -
**dave**, Sunday, November 29, 2009 at 4:14pmthank you very much.. i understand what you are saying, but can you show me the algebra steps to get the 2 solutions?

i keep substituting the first equation into the second and try to work it out, but for some reason i am still doing it wrong.

- college physics -
**MathMate**, Sunday, November 29, 2009 at 4:17pmIt's time you show me your work!

I will be pleased to help you find the problem if there is any.

- college physics -
**dave**, Sunday, November 29, 2009 at 6:51pmmy variables are a little different than yours, but here is what i have..

Yf=Vosinθ (tf)-.5g(tf)²

Yf=Vosinθ(xf)/Vocosθ-.5g(xf/Vocosθ)²

i'm having trouble isolating Vo. am i supposed to use the quadratic equation? my algebra is not very good.. so if you can point me in the right direction it would help alot.

- college physics -
**MathMate**, Sunday, November 29, 2009 at 8:57pmFortunately there is no quadratic equation to solve, just a square-root.

Yf=Vosinθ(xf)/Vocosθ-.5g(xf/Vocosθ)²

Yf- sinθ(xf)/ cosθ=.5g(xf/Vocosθ)²

Vo²

=0.5g(xf/cosθ)²/(Yf-(xf)tanθ), or

Vo=sqrt(0.5g(xf/cosθ)²/(Yf-(xf)tanθ))

Now take out your calculator and substitute the values. You should get about 46.6 m/s.

- college physics -
**dave**, Sunday, November 29, 2009 at 9:30pmahhh i finally got it!!! thank you so much!

- college physics -
**MathMate**, Sunday, November 29, 2009 at 10:08pmHow much did you get?

- college physics -
**MathMate**, Sunday, November 29, 2009 at 10:08pmHow much did you get?

- college physics -
**MathMate**, Sunday, November 29, 2009 at 10:09pmWhat velocity did you get?

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