How is a projectile’s initial speed related to its height and horizontal displacement?
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OpenStudy (billyjean):
@paki
Parth (parthkohli):
Well, what do you know about projectiles to begin with?
OpenStudy (billyjean):
I only know that a projectile is a missile designed to be fired from a rocket or gun.
Parth (parthkohli):
So the math part of it?
OpenStudy (billyjean):
i dont know im confused
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Parth (parthkohli):
Well, alright. So do you know equations of motion?
OpenStudy (billyjean):
no
Parth (parthkohli):
Aw man, I can't help you out then
OpenStudy (billyjean):
why
Parth (parthkohli):
you need to be familiar with the basics to figure this out
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OpenStudy (billyjean):
v = u + at From acceleration definition
s = ½ (v+u) × t Average velocity is ½ (v+u)
s = ut + ½ at² No 'v' needed
v² = u² + 2as No 't' needed
OpenStudy (billyjean):
is that it
OpenStudy (billyjean):
@3mar
OpenStudy (3mar):
Oh
I was checking that question before and have prepared something for it, but it was closed!
OpenStudy (bugman954):
Simple the greater the speed the greater the height and horizontal displacement
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OpenStudy (vipulrakta):
|dw:1480567921789:dw| and the relation between u; h , r is\[H = (u ^{2}\sin ^{2} \theta) / 2g , R=(u^{2} \sin 2\theta )/g\] grater the initial speed more is horizantal distance and hight is more ance vice versa . R is maximum when angle = 45