Projectile Motion Q (1 Viewer)

[shaon0]

The reason you didn't get this question is because you don't know the equations of the paths of motion.

Look in your textbook and follow the exercises which teach you:

y'' = -g
y' = -gt + Vsin[a]
y = -(1/2)gt^2 + Vt sin[a]

x'' = 0
x' = V cos[a]
x = Vtcos[a]

Once you know these equations of motion, you can solve these problems quite easily.

Edit: Didn't realise there was a second page, everything I said has already been said, but it bears repeating that you should be able to prove these equations of motion to yourself, and to the examiner.

Its on the page after the exercise i'm doing.

 

[tacogym27101990]

I'm doing them right now. but they are more like physics then math. They don't require you to use calculus but instead use formulae.

you arent allowed to use the physics formulas in maths
you have to derive all the equations

 

[shaon0]

you arent allowed to use the physics formulas in maths
you have to derive all the equations

Oh okay. didn't know that.

 

[shaon0]

Hey, does anyone know where to get projectile motion resources from for 3unit?

 

[shaon0]

Question:
A bullet is fired horizontally with a velocity of 800m/s from a height of 3m above the horizontal plane. What is the range of the bullet?
I don't understand the horizontally part. Do they mean at 0 degrees?

 

[u-borat]

yep.

 

[shaon0]

yeah,

means u_x = 800, u_y = 0

how would i find the range?

 

[u-borat]

depending on how you set up your original equations, you put the y height of the projectile equal to zero or -3.

 

[3unitz]

Question:
A bullet is fired horizontally with a velocity of 800m/s from a height of 3m above the horizontal plane. What is the range of the bullet?
I don't understand the horizontally part. Do they mean at 0 degrees?

derive the two equations:

S_y = u_yt - 9.8 t^2/2 -----(1)
S_x = u_xt -----(2)

from (1):
- 3 = - 9.8 t^2/2
t ~ 0.78sec

sub into (2):
R = 800 x 0.78
~ 626m

 

[shaon0]

derive the two equations:

S_y = u_yt - 9.8 t^2/2 -----(1)
S_x = u_xt -----(2)

from (1):
- 3 = - 9.8 t^2/2
t ~ 0.78sec

sub into (2):
R = 800 x 0.78
~ 626m

I tried that but i think i made a mistake. I took out the parameter and put it in cartesian form.

 

[JamesTockuss]

Let Uy = 45sinX*

then v = u + at
0 = 45sinx* + (-9.8)(4)