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?

 

[DeathEater]

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.

I used the equation for time to fall which is 2.622 = √(2*h/g)

 

[DeathEater]

Ok.

Ok.

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

Well how would I find the launch angle then?

 

[tnich]

I used the equation for time to fall which is 2.622 = √(2*h/g)

You say that t1 = 2.473s, but that is not the same value you had for it initially (2.34s).

 

[tnich]

Well how would I find the launch angle then?

Once again, I think it would really help to draw a diagram of the trajectory and label it with what you already know. Then you could start figuring out how to find the launch angle.

 

[haruspex]

Well how would I find the launch angle then?

What equations relate the initial velocity components?

 

[DeathEater]

What equations relate the initial velocity components?

I don't know
I'm confused as to why you can't just use inverse tan

 

[DeathEater]

Once again, I think it would really help to draw a diagram of the trajectory and label it with what you already know. Then you could start figuring out how to find the launch angle.

I have and I still don't know

 

[DeathEater]

You say that t1 = 2.473s, but that is not the same value you had for it initially (2.34s).

I reworked it again and got 2.34, my bad. Is that even correct?

 

[tnich]

I reworked it again and got 2.34, my bad. Is that even correct?

Yes. Now you need to figure out how to get initial velocity.

 

[DeathEater]

Yes. Now you need to figure out how to get initial velocity.

well vi*cosθ * (2.477*2) = 53 gives that vi = 10.698/cosθ
and could I use the information about the max height?

doing that I end up with: 2.477= (vi*sinθ)/g multiply both sides by g and get that 24.3/sinθ = vi
setting the two equations equal : 10.698/cosθ = 24.3/sinθ and get tan^-1(24.3/10.698) = 66.239°

plugging that back in: vi= 10.698/ cos(66.239)= 26.55 m/s?

 

[tnich]

well vi*cosθ * (2.477*2) = 53 gives that vi = 10.698/cosθ
and could I use the information about the max height?

doing that I end up with: 2.477= (vi*sinθ)/g multiply both sides by g and get that 24.3/sinθ = vi
setting the two equations equal : 10.698/cosθ = 24.3/sinθ and get tan^-1(24.3/10.698) = 66.239°

plugging that back in: vi= 10.698/ cos(66.239)= 26.55 m/s?

I think you've got it.

The way you have done it is perfectly fine. I want to point out another way that you might find useful in other problems. You started by writing equations for the horizontal and vertical components of the initial velocity given the time t at which the body reaches point (25, 30). Then you solved for t. Now you can substitute the value of t back into those equations to get the initial velocity components. Then you can use those components to get the magnitude and direction (θ and vi) of the velocity vector.
More generally, when you have two equations with sines and cosines, you can often square and add them to eliminate the angle or divide them to get the tangent of the angle. In this case

v t cos(θ) = x
v t sin(θ) = y + ½ g t²

v² t² cos²(θ) = x²
v² t² sin²(θ) = (y + ½ g t²)²

v² t² = x² + (y + ½ g t²)²
tan(θ) = (y + ½ g t²)/x

 

[haruspex]

You started by writing equations for the horizontal and vertical components of the initial velocity given the time t at which the body reaches point (25, 30). Then you solved for t.

I have been following this thread, somewhat intrigued by the approach of solving for t first. It is unexpected because we do not in the end care about the time to clear the obstacle.
In fact, nowhere in the thread do I see how this solving for t is done. It cannot be purely based on the two equations you mention since they include two other unknowns, v and θ, and different combinations of those lead to differenr values of t. So I have to assume that the range information was also used.

The more obvious approach, to me, is to eliminate t from the two equations, then eliminate v² using the range equation. Does that turn out to be more complicated? Without seeing your method I cannot be sure, so I post mine for comparison.

Let x, y be the coordinates of the (25, 30) point and r be the range. For typing convenience I will abbreviate sin, cos and tan of the angle to s, c and τ.
1. y=v s t - ½gt²
2. x=v c t
3. gr=2v²sc
From 1 and 2, $2v^2c^2=\frac{gx^2}{x\tau-y}$
From 3, $gr=\tau\frac{gx^2}{x\tau-y}$
$rx\tau-ry=\tau x^2$
$\tau=\frac{ry}{x(r-x)}$
We can see that this has the correct behaviour as x→0 and as x→r. The trajectory approaches the vertical.

 

[tnich]

I have been following this thread, somewhat intrigued by the approach of solving for t first. It is unexpected because we do not in the end care about the time to clear the obstacle.
In fact, nowhere in the thread do I see how this solving for t is done. It cannot be purely based on the two equations you mention since they include two other unknowns, v and θ, and different combinations of those lead to differenr values of t. So I have to assume that the range information was also used.

The more obvious approach, to me, is to eliminate t from the two equations, then eliminate v² using the range equation. Does that turn out to be more complicated? Without seeing your method I cannot be sure, so I post mine for comparison.

Let x, y be the coordinates of the (25, 30) point and r be the range. For typing convenience I will abbreviate sin, cos and tan of the angle to s, c and τ.
1. y=v s t - ½gt²
2. x=v c t
3. gr=2v²sc
From 1 and 2, $2v^2c^2=\frac{gx^2}{x\tau-y}$
From 3, $gr=\tau\frac{gx^2}{x\tau-y}$
$rx\tau-ry=\tau x^2$
$\tau=\frac{ry}{x(r-x)}$
We can see that this has the correct behaviour as x→0 and as x→r. The trajectory approaches the vertical.

What I meant was, write the range equation as

1. 2 v sin(θ) v cos(θ) = r g

write the vertical and horizontal displacement equations as

2. v sin(θ) = (y +½ g t²)/t
3. v cos(θ) = x/t

then substitute 2. and 3. into 1. and solve for t. Then use t to solve 2. and 3. for horizontal and vertical components of velocity.

 

[haruspex]

What I meant was, write the range equation as

1. 2 v sin(θ) v cos(θ) = r g

write the vertical and horizontal displacement equations as

2. v sin(θ) = (y +½ g t²)/t
3. v cos(θ) = x/t

then substitute 2. and 3. into 1. and solve for t. Then use t to solve 2. and 3. for horizontal and vertical components of velocity.

Ok, thanks for the clarification.
The simplest along those lines would be to solve the resulting linear equation in t² (no need to find t, as such) and substitute that into the expression for tan θ had from dividing (2) by (3) to get the expression I got in post #42.

Looks about equal to me.

 

[tnich]

Ok, thanks for the clarification.
The simplest along those lines would be to solve the resulting linear equation in t² (no need to find t, as such) and substitute that into the expression for tan θ had from dividing (2) by (3) to get the expression I got in post #42.

Looks about equal to me.

Right. I also like that you can square and add 1. and 2. and then solve for v in terms of t².