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1. Fairly easy, however I have forgotten all this stuff

A coastal defence cannon, positioned at the top of a 150m vertical cliff, is able to fire shells at a velocity of 120ms. If fired horizontally out from the cliff, how far out to sea will the shell land?

(b) The cannon is now aimed up at an angle of 45 degrees above horizontal. How far will a shell now travel?
2. (a)vertically,
u= 0 a= 9.8 x= 150

x= ut + 0.5at^2

work out t then subs horizontally into x= ut where u=120 to find x.

(b) vertically, u is now u= 120sin45 and horizontally u = 120cos 45 (they're actually the same). use same method, find t then horizontal distance.
3. a=-9.8
s=-150
u=0
t= ?
s = ut +1/2 at^2
-150 = -4.9t^2
t=5.5s
s= ?
u = 120
a = 0
s = ut + 0.5at^2
s = 663m

Same again except
u = 120sin45 for first part
and u = 120cos45 for second
4. ok im doing the first part for part b, using your methods. You get a quadratic, however when you try to solve you get two values for t. Which value do you use?
5. Sorry should have been clearer on the second one.

If you take up as positive, then vertically
u= 120sin45 a=-9.8 (opposing) x= -150
(sin45= 1/√2 )

-150= 120t/√2 - 4.9t²
4.9t² -120t/√2 -150=0

If you solve that you get one negative and on positive value, obviously t can only be positive.
6. I get 137.2m, does that sound right?
7. nope. think about it, if you fire it up at an angle, it will travel even further than before.

solving 4.9t² -120t/√2 -150=0
i get t= 18.93s

then distance horizontally,
x= 18.93*120/√2
x= 1607m
8. Yeah I was thinking about that. Thanks for the help, I missed a - in the quadratic formula which probably threw it off (dont do physics at midnight).

A rifle is fired up at an angle of 55 degrees above horizontal. If the initial velocity of the bullet is 570ms, what wil lbe its velocity (magnitude and direction) 2.5 s after firing?

x component = 570cos55
y component = 570sin 55

So I guess we just find delta y, and then use that in the equation (v^2 = u^2 + 2a(delta y)) ?
10. (Original post by bally)

A rifle is fired up at an angle of 55 degrees above horizontal. If the initial velocity of the bullet is 570ms, what wil lbe its velocity (magnitude and direction) 2.5 s after firing?

x component = 570cos55
y component = 570sin 55

So I guess we just find delta y, and then use that in the equation (v^2 = u^2 + 2a(delta y)) ?
It's been a while since A'level ...

v = u + at

.: Vy = (570sin55) + (-9.8)(2.5)
.: Vy = 442.4166652 ms^-1

=> V = sqrt[(442.4166652)^2 + (570cos55)^2]
.: V = 550.1 ms^-1

=> theta = arctan[442.4166652 / (570cos55)]
.: theta = 53.5

So, velocity is 550.1 ms-1 at 53.5o to the horizontal.
11. (Original post by bally)
thanks, but i dont get where this equation comes from?

=> V = sqrt[(442.4166652)^2 + (570cos55)^2]
It's just Pythagoras. You have the new y component and the x component and need to find the resultant...
12. edit; lol just realised in time (about pythagoras)

=> theta = arctan[442.4166652 / (570cos55)]
.: theta = 53.5

Isnt the 442.... the hypotenuse and the 570cos55 the bottom of the triangle? How does that work?

edit: oh wait this is the new triangle where we resolve the forces to find the resultant angle...
13. (Original post by bally)
edit; lol just realised in time (about pythagoras)

=> theta = arctan[442.4166652 / (570cos55)]
.: theta = 53.5

Isnt the 442.... the hypotenuse and the 570cos55 the bottom of the triangle? How does that work?
The 442 is the vertical (side) and 570cos55 is the horizontal (bottom).

We know that: tan theta = opposite / adjacent

and so therefore theta = arctan[opposite/adjacent]. Arctan is just the inverse tan function, sometime written as tan-1
14. (Original post by bally)
ed
edit: oh wait this is the new triangle where we resolve the forces to find the resultant angle...
yep

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