# Thread: Projectile Motion :D

1. ## Projectile Motion :D

A horizontal drainpipe 6m above sea level empties stormwater into the sea. If the water comes out horizontally and reaches the sea 2m out from the pipe, find the initial velocity of the water, correct to 1 decimal place. let g = 10ms^-2 and neglect air resistance.

2. ## Re: Projectile Motion :D

the angle is 90 because it is launched horizontally.

enjoy

3. ## Re: Projectile Motion :D

Originally Posted by untouchablecuz
the angle is 90 because it is launched horizontally.

enjoy
Um... doesn't horizontally mean the angle is 0?

and x(dot) should be vcos0? y(dot) should be vsin0? (when t = 0)

4. ## Re: Projectile Motion :D

Originally Posted by Michaelmoo
Um... doesn't horizontally mean the angle is 0?
yeah thats right, not 90

5. ## Re: Projectile Motion :D

Originally Posted by clintmyster
yeah thats right, not 90
Wait maybe he takes his angle relative to the vertical. Bit wierd though...

6. ## Re: Projectile Motion :D

Originally Posted by Michaelmoo
Um... doesn't horizontally mean the angle is 0?

and x(dot) should be vcos0? y(dot) should be vsin0? (when t = 0)
for these Q's, it take it from the vertical

and notice that y(dot) = vsin0 = vcos(90-0) = vcos90 and the same with x(dot)

7. ## Re: Projectile Motion :D

so should i be using the angle of projection = 0 or 90? :S

8. ## Re: Projectile Motion :D

Originally Posted by Smilebuffalo
so should i be using the angle of projection = 0 or 90? :S
works with both, bit what would be the angle if something is 'dropped' vertically down from a height?

9. ## Re: Projectile Motion :D

no angle, all forces are already in the same vertical plane

there is only a gravitational force down (ignoring air resistance)

i.e F=mg => y(double dot) = g

If wind resistance is taken into account,

y(double dot) =g-R/m

10. ## Re: Projectile Motion :D

can someone please show me the entire method of solving this question using the angle = 0.

11. ## Re: Projectile Motion :D

$x'' = 0\\ x' = v0\\ x = v0t \to \text{solve for 2m, so } 2 = v0t \to v0 = 2/t\\ y'' = g \text{ Note, taking down as positive direction.}\\ y' = gt\\ y = \frac{gt^2}{2} \text{ It drops 6metres, so } 12 = gt^2\\ t^2 = 1.2 \to t = \sqrt{1.2} \\ \text{Subbing this into v0 equation } \to v0 = \frac{2}{\sqrt{1.2}} \\ \text{Thus } v0 = 1.8257$

12. ## Re: Projectile Motion :D

Originally Posted by untouchablecuz
for these Q's, it take it from the vertical

and notice that y(dot) = vsin0 = vcos(90-0) = vcos90 and the same with x(dot)
Just curious. Any particular reaason you take it relative to the vertical?

13. ## Re: Projectile Motion :D

Originally Posted by Michaelmoo
Just curious. Any particular reaason you take it relative to the vertical?
no patricular reason, it seems more logical

i've taught my self >90% of the 3 unit course and this has always been the way i've done it

tbh, i always thought everyone did it like me

14. ## Re: Projectile Motion :D

mate, when t=0,y=6, so y does not equal to gt^2/2, also you forgot to +c

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