Calculating Projectile Height: 30 Degrees vs. 60 Degrees

[HadanIdea]

Hi,

How does one prove that a projectile will shoot 3x higher at 60 degrees that at 30 degrees?

And there is no more info available.

Here is what i attempted:

I gave it 500m/s,

1a.) The formula for maximum time @ 60 degrees: th = (u)(Sin60)/9.8
= (500)(Sin60)/9.8
= 44.185's

2a.) The formula for maximum height @ 60 degrees: h = (u)(t) + (1/2)(a)[(t)"squared"]
= (500)(44.185) + (1/2)(9.8)[(44.185)"squared"]
= 31658.8397m (31.659km)

1b.) The formula for maximum time @ 30 degrees: th = (u)(Sin30)/9.8
= (500)(Sin30)/9.8
= 25.510's

2b.) The formula for maximum height @ 30 degrees: h = (u)(t) + (1/2)(a)[(t)"squared"]
= (500)(25.510) + (1/2)(9.8)[(25.510)"squared"]
= 15943m (15.944km)

3a.) Now if i divide height of 60 degrees with the height of 30 degrees
I get 31658.8397m / 15943m = 1.986 times the height and not 3x the height

Did i do it right or am i looking at it in a wrong way?

Regards

 

[dreamLord]

Your method is fine. It must be the calculations. I would suggest doing it with variables instead of specific values - that way you don't make errors calculating.

EDIT : I noticed you are using value of u as 500m/s in the distance equations. That is incorrect - you must use the upward value of velocity, which is u*sin*theta (theta = initial angle).

 

[tiny-tim]

Welcome to PF!

Hi HadanIdea! Welcome to PF!

Nobody asked you for t, so why are you bothering to find it??

Use another constant acceleration equation, that doesn't involve t !

 

[SteamKing]

In equations 2a. and 2b. in the OP, you use the initial velocity of 500 m/s in calculating the max. height of the projectile. This is incorrect. You must use only the vertical component of the initial velocity in calculating the max. height of the projectile.

 

[Solkar]

Don't put in sample numbers at an early stage; look at the big picture first and find proportionalities.

Let "_v" denote vertical, "_h" denote horizontal components, "∝" means "is proportional to"

Decompose momentum
p =: p_v + p_h,

consider Energies
V ~ mgh => V ∝ h
T = p²/2m

and think about
- if g affects p_h
- how to decompose T into T_v and T_h
- and think about p_v and T_v at the top of the trajectory.

 

[TSny]

HadanIdea,

My 2 cents advice is to heed tiny-tim's 2 dollar advice!

 

[PeterO]

Hi,

How does one prove that a projectile will shoot 3x higher at 60 degrees that at 30 degrees?

And there is no more info available.

Here is what i attempted:

I gave it 500m/s,

1a.) The formula for maximum time @ 60 degrees: th = (u)(Sin60)/9.8
= (500)(Sin60)/9.8
= 44.185's

2a.) The formula for maximum height @ 60 degrees: h = (u)(t) + (1/2)(a)[(t)"squared"]
= (500)(44.185) + (1/2)(9.8)[(44.185)"squared"]
= 31658.8397m (31.659km)

1b.) The formula for maximum time @ 30 degrees: th = (u)(Sin30)/9.8
= (500)(Sin30)/9.8
= 25.510's

2b.) The formula for maximum height @ 30 degrees: h = (u)(t) + (1/2)(a)[(t)"squared"]
= (500)(25.510) + (1/2)(9.8)[(25.510)"squared"]
= 15943m (15.944km)

3a.) Now if i divide height of 60 degrees with the height of 30 degrees
I get 31658.8397m / 15943m = 1.986 times the height and not 3x the height

Did i do it right or am i looking at it in a wrong way?

Regards

When resolving the 30 degree - 60 degree launch angles, you will be resolving the launch speed using the familiar 2-1-√3 triangle.

This shows that , compared to a 30 degree launch, the vertical speed is larger by a factor of √3.

That means the average vertical velocity before reaching maximum height is higher by a factor of √3.
The time taken to slow to 0 vertical velocity is also up by a factor of √3

So you travel at √3 x the speed for √3 x the time.

What result will that give?

 

[TSny]

When resolving the 30 degree - 60 degree launch angles, you will be resolving the launch speed using the familiar 2-1-√3 triangle.

This shows that , compared to a 30 degree launch, the vertical speed is larger by a factor of √3.

That means the average vertical velocity before reaching maximum height is higher by a factor of √3.
The time taken to slow to 0 vertical velocity is also up by a factor of √3

So you travel at √3 x the speed for √3 x the time.

What result will that give?

Nice.