- #1
HadanIdea
- 6
- 0
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
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