Initial velocity if a performer clears a 30 m tall object

[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.