Initial velocity if a performer clears a 30 m tall object

[DeathEater]

Homework Statement

In 1940, Emanuel Zachinni, one of a family of “human cannonball” performers, set a world record by traveling 53m. If he clears the 20 m tall ferris wheel (which is 25 m away from the cannon) by 10 m, how fast was he going at launch?

Homework Equations

range 53m =[(vi)²sin(2θ)]/g
time vi*cos(θ)*t = 25m & vi*sin(θ)*t -½*g*t² = 30 m
(since this occurs at same time):
25/ vi*cosθ = t
which can be plugged into vi*sin(θ)*t -½*g*t2 = 30 m

The Attempt at a Solution

tried solving for vi to plug in using range and got √((53*g)/sin(2θ)) . I don't really now where to go from here. :/

 

[haruspex]

25/ vi*cosθ = (-60+vi*sinθ)/g

How do you get the right hand side of that equation?

 

[tnich]

You have three equations in three unknowns, so you need to start by combining them in some way to eliminate all but one variable. You have made a start at that, but you have incorrectly solved the vertical speed equation for t. Since that equation is quadratic in t, you are going to end up with vi*sin(θ) under a radical, which will not be very useful. You could try instead to eliminate vi²sin(2θ) from the range equation, replacing it with an expression involving t.

 

[DeathEater]

How do you get the right hand side of that equation?

I see my mistake now. Should it be (1/2)gt² - vi*sinθ*t - 30?

and do I use the quadratic formula for t?

 

[haruspex]

vi*sin(θ)*t -½*g*t2 = 30 m by solving for t using this

That would have given you something with a square root in it. What happened to that?

Try the other way, use the horizontal equation to get a t value to substitute in the vertical equation.

 

[DeathEater]

That would have given you something with a square root in it. What happened to that?

Try the other way, use the horizontal equation to get a t value to substitute in the vertical equation.

are you saying to plug in 25/vi*cosθ for t in the expression (1/2)gt2 - vi*sinθ*t - 30? Because I'm not sure what to do with the range equation since the t's are different.

 

[DeathEater]

You have three equations in three unknowns, so you need to start by combining them in some way to eliminate all but one variable. You have made a start at that, but you have incorrectly solved the vertical speed equation for t. Since that equation is quadratic in t, you are going to end up with vi*sin(θ) under a radical, which will not be very useful. You could try instead to eliminate vi²sin(2θ) from the range equation, replacing it with an expression involving t.

I'm sorry, I am not quite sure how I could eliminate that since the t's are different in the 3 equations

 

[haruspex]

are you saying to plug in 25/vi*cosθ for t in the expression (1/2)gt² - vi*sinθ*t - 30?

Yes. (I assume you meant (1/2)gt² - vi*sinθ*t = 30.)

I'm not sure what to do with the range equation since the t's are different.

You range equation does not feature a time.

 

[tnich]

I'm sorry, I am not quite sure how I could eliminate that since the t's are different in the 3 equations

I am not sure why you say that. Didn't you set up the two time equations to represent the vertical and horizontal positions at the same time t? Also, t does not appear at all in the range equation.

 

[DeathEater]

Yes. (I assume you meant (1/2)gt² - vi*sinθ*t = 30.)

You range equation does not feature a time.

okay, I tried doing that and ended up with:
3065.625/(vi²*cos²θ) - 25 tanθ = 30

do I solve for vi² using that and plug it into range to solve for θ?

the 3065.625 also comes from the 25²*½*g

 

[DeathEater]

I am not sure why you say that. Didn't you set up the two time equations to represent the vertical and horizontal positions at the same time t? Also, t does not appear at all in the range equation.

yes I see the issue (see my newest comment above). The math just seems a bit "ugly" for 1st year physics so I assumed I was doing something wrong.

 

[tnich]

yes I see the issue (see my newest comment above). The math just seems a bit "ugly" for 1st year physics so I assumed I was doing something wrong.

There is a trick that makes it pretty easy. See my first comment.

 

[DeathEater]

There is a trick that makes it pretty easy. See my first comment.

yeah I see it, I just don't understand how to get rid of the vi²sin(2θ). You say to replace it with a t, but I don't know what you mean by that. I know the range equation is t(final) * vi*cosθ, but the t's in the other equations aren't the same

 

[haruspex]

do I solve for vi2 using that and plug it into range to solve for θ?

Seems a reasonable strategy.

 

[tnich]

yeah I see it, I just don't understand how to get rid of the vi²sin(2θ). You say to replace it with a t, but I don't know what you mean by that

Write sin(2θ) in terms of sinθ and cosθ. Then find a way to express sinθ and cosθ in terms of t.

 

[DeathEater]

Seems a reasonable strategy.

well when I tried it I got θ= 116.12°...so.

 

[DeathEater]

Write sin(2θ) in terms of sinθ and cosθ. Then find a way to express sinθ and cosθ in terms of t.

do you mean re-writing vi*cos(θ)*t = 25m & vi*sin(θ)*t -½*g*t2 = 30 m to find cosθ and sinθ and plug that in?

 

[tnich]

do you mean re-writing vi*cos(θ)*t = 25m & vi*sin(θ)*t -½*g*t2 = 30 m to find cosθ and sinθ and plug that in?

 

[tnich]

That does seem like a way to write sinθ and cosθ in terms of t.

 

[DeathEater]

That does seem like a way to write sinθ and cosθ in terms of t.

but I still end up with vi and t, so 2 variables? what do I do about that?

 

[tnich]

but I still end up with vi and t, so 2 variables? what do I do about that?

Work it through and see what you get.

 

[DeathEater]

Work it through and see what you get.

I see my mistake. I am an idiot haha.

I got t = 2.34 s (approx)

 

[DeathEater]

I see my mistake. I am an idiot haha.

I got t = 2.34 s (approx)

but the θ came out as -7.33°?

 

[tnich]

but the θ came out as -7.33°?

Why are you calculating θ? I thought you wanted vi.

 

[tnich]

Why are you calculating θ? I thought you wanted vi.

How did you calculate θ?

 

[DeathEater]

How did you calculate θ?

took cos^-1(((125/53)-5)/6) and after recalculating got 116°. Can you help me figure out where I am going wrong?

 

[tnich]

took cos^-1(((125/53)-5)/6) and after recalculating got 116°. Can you help me figure out where I am going wrong?

I can't tell how you got there. Since you have cos^-1 in your expression, I assume you are substituting values for t and vi into the horizontal position equation. What value are you using for vi?

 

[DeathEater]

I can't tell how you got there. Since you have cos^-1 in your expression, I assume you are substituting values for t and vi into the horizontal position equation. What value are you using for vi?

Okay, I tried a different approach. I know that at t1= 2.473 s, the man has a horizontal displacement of 25 m. Since the peak will be reached when he is 26.5 m away, I set up an equality expression and got that t2 (time at peak) is 2.622 s

using the formula for time to fall, I found that the peak height is 33.71 m
using tan^-1(33.71/26.5) I got θ= 51.827°

and since vi*cosθ*time final = 53
then vi = 53/ (cos(51.827)*(2.622*2)) = 16.353 m/s

Is this now correct?

 

[tnich]

Okay, I tried a different approach. I know that at t1= 2.473 s, the man has a horizontal displacement of 25 m. Since the peak will be reached when he is 26.5 m away, I set up an equality expression and got that t2 (time at peak) is 2.622 s

using the formula for time to fall, I found that the peak height is 33.71 m
using tan^-1(33.71/26.5) I got θ= 51.827°

and since vi*cosθ*time final = 53
then vi = 53/ (cos(51.827)*(2.622*2)) = 16.353 m/s

Is this now correct?

Your conceptual approach is a good one, but I think your value for the peak height is incorrect. Since you haven't shown how you calculated it, I can't say where your error is. Also, I suggest that you draw a diagram of the trajectory to see if the angle you are calculating is the one you really want.

 

[haruspex]

the peak will be reached when he is 26.5 m away,

Ok.

the peak height is 33.71 m

Ok.

using tan-1(33.71/26.5) I got θ= 51.827°

What has that got to do with the launch angle? Was the trajectory a straight line up to the peak?