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Q on projectiles (1 Viewer)
- Thread starter jmcloon
- Start date
Right well on the formula sheet you will have these equations:jmcloon said:
v = u + at
v<sup>2</sup> = u<sup>2</sup> + 2as [s or r depending on what you want as displacement]
s = ut + 1/2 a t<sup>2</sup>
The horizontal velocity will always be constant hence v<sub>x</sub> = u<sub>x</sub>
and a<sub>x</sub> = 0
Working vertically,
a<sub>y</sub> = -9.8 [Assuming earth, negative due to towards center of planet]
I personally use the following table:
---x------y
r
u
v
a
t
Filling these out with what you know, look for what you need to find out and use one of above formulae
i find that the best way is to use galileo's method (which is the method we're taught):
first, you've gotta split the horizontal and vertical components up using a right angled triangle, the initial velocity, and the angle (which you should also get given). the best way is to draw the triangle on the page and then apply SOH CAH to it (sine(angle)=opposite(vertical)/hypotenuse, Cos(angle) = adjacent(horizontal)/hypotenuse) . you now have your component velocities
to work out the max height, you use your vertical component and v^2 = u^2 + 2as. u=your vertical velocity, v=0 (at the top of the arc), a = g = 9.8. just reearrange and you've got your answer!
trip time, you need to use s=ut + 1/2 *at^2, s=0(or height back on the ground), u = the vertical velocity, a = 9.8. the trick with this one is the factorising of it, because it is a t^2 there are two answers, 0, and the one you are looking for. what you need to do is to take the t out so you have 0 = t(u + 4.9t) so t now equals 0 or u/4.9. sometimes, when s <> 0 you have to use the quadratic formula, (the one from maths).
to find range you just need s = ut:
you need to use the trip time to find this out. u = horizontal velocity, t=trip time, sub them in and thats your answer
the trick to projectile motion is to treat horizontal and vertical components seperately the only thing that really relates them is time.
hope that helps
first, you've gotta split the horizontal and vertical components up using a right angled triangle, the initial velocity, and the angle (which you should also get given). the best way is to draw the triangle on the page and then apply SOH CAH to it (sine(angle)=opposite(vertical)/hypotenuse, Cos(angle) = adjacent(horizontal)/hypotenuse) . you now have your component velocities
to work out the max height, you use your vertical component and v^2 = u^2 + 2as. u=your vertical velocity, v=0 (at the top of the arc), a = g = 9.8. just reearrange and you've got your answer!
trip time, you need to use s=ut + 1/2 *at^2, s=0(or height back on the ground), u = the vertical velocity, a = 9.8. the trick with this one is the factorising of it, because it is a t^2 there are two answers, 0, and the one you are looking for. what you need to do is to take the t out so you have 0 = t(u + 4.9t) so t now equals 0 or u/4.9. sometimes, when s <> 0 you have to use the quadratic formula, (the one from maths).
to find range you just need s = ut:
you need to use the trip time to find this out. u = horizontal velocity, t=trip time, sub them in and thats your answer
the trick to projectile motion is to treat horizontal and vertical components seperately the only thing that really relates them is time.
hope that helps