Finding Horizontal Displacement for Spring Cannon Launch

[monac]

A spring cannon is located at the edge of a table that is 1.2 m above the floor. A steel ball is launched from the cannon with speed Vo at 35.0 degrees above the horizontal.
(a) Find the horizontal displacement component of the ball tothe point where it lands on the floor as a function of Vo. We writethis function as x(Vo).
Evaluate x for (b) Vo = 0.100 m/s and for
(c) Vo = m/s.
(d) Assume Vo is close to zero but not equal to zero.Show that one term in the answer to part (a) dominates so that thefunction x(Vo) reduces to a simpler form.
(e) If Vo is ver large,what is the approximate form of x(Vo).

I found the answers to a, b, and c. I am thinking of taking a limit on d but the problem is really complicated ... It's a quadratic and with limit laws I would set it equal to 0, but it says it can't be 0. So i am really confused on how to do d and e.

 

[Curious3141]

A spring cannon is located at the edge of a table that is 1.2 m above the floor. A steel ball is launched from the cannon with speed Vo at 35.0 degrees above the horizontal.
(a) Find the horizontal displacement component of the ball tothe point where it lands on the floor as a function of Vo. We writethis function as x(Vo).
Evaluate x for (b) Vo = 0.100 m/s and for
(c) Vo = m/s.
(d) Assume Vo is close to zero but not equal to zero.Show that one term in the answer to part (a) dominates so that thefunction x(Vo) reduces to a simpler form.
(e) If Vo is ver large,what is the approximate form of x(Vo).

I found the answers to a, b, and c. I am thinking of taking a limit on d but the problem is really complicated ... It's a quadratic and with limit laws I would set it equal to 0, but it says it can't be 0. So i am really confused on how to do d and e.

Why don't you show what you've done so far (to all parts)?

Just a hint, but it'll likely be helpful. Remember the Taylor series (or Binomial theorem) for things like (1+x)^-1.

 

[monac]

it's a long process. I ll just type the main parts of the answers.

(a) x = vicos35t
t = (vi sin35 + √vi^2 sin^2 35 + 23.544) / 9.81
so x = vi cos35 ((vi sin35 + √vi^2 sin^2 35 + 23.544) / 9.81)

for b and c i just had to plug in values for vi and get the x so that was easy.

I have not learned Taylor's series yet in my Calculus II class. :(

 

[Curious3141]

it's a long process. I ll just type the main parts of the answers.

(a) x = vicos35t
t = (vi sin35 + √vi^2 sin^2 35 + 23.544) / 9.81
so x = vi cos35 ((vi sin35 + √vi^2 sin^2 35 + 23.544) / 9.81)

for b and c i just had to plug in values for vi and get the x so that was easy.

I have not learned Taylor's series yet in my Calculus II class. :(

Sorry, your expression is ambiguous. √vi^2 = vi, isn't it?

 

[monac]

Oh the whole thing is under the square root.
t = (vi sin35 + √(vi^2 sin^2 35 + 23.544)) / 9.81

 

[NascentOxygen]

I found the answers to a, b, and c. I am thinking of taking a limit on d but the problem is really complicated ... It's a quadratic

I don't think you need any more maths for (d). Just inspect the expression you earlier derived for time but now saying that anything involving Vo is insignificant compared with associated other terms.

 

[monac]

So imagine that vi = 0?

 

[NascentOxygen]

So imagine that vi = 0?

For (d), close to 0. (Careful with the notation. The question uses Vo and you seem to have changed it to vi.)

 

[monac]

In the back of the book, the answer for d was x ~ 0.405 m

 

[NascentOxygen]

In the back of the book, the answer for d was x ~ 0.405 m

Regardless of the speed it's launched, providing it's slow it still travels 0.4m horizontally? That's interesting.

My answer for (d) is X_H = þVo
where þ is a fraction that I'll leave you to work out for yourself, but it's between 0.1 and 0.9