Physics question- projectile motion (1 Viewer)

elfanger

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Need hlep with this question, its been annoying me for a few days now. It has a diagram attached to it so i drew it up in paint.
It is attached.

Thanks.
 

perfectionist

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LOL i was just looking at the Question...sorry...i forgot how to do it :(...anyways...i don't need it any more :p
 

speersy

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just work out the horizontal component of velocity using simple trigonometry. actually now i think about it i cant explain it in words i need to do calculations which would take too much time to write using a keyboard. sorry

anyway it will do wonders for your confidence and problem solving skills if you perservere and do it yourself.
 

+:: $i[Q]u3 ::+

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Originally posted by speersy
anyway it will do wonders for your confidence and problem solving skills if you perservere and do it yourself.
quite true.. =) but we figure the question's been put up and we should answer it...

in y-direction (origin is at cliff level, positive direction upwards).
s = ut + (1/2)at^2
-50 = 60sin45 x t + (1/2) x (-9.8) x t^2
(rearranging)
(4.9)t^2 - (60sin45)t - 50 = 0
you have a quadratic in t, which u can sove.
(you wil end up with two values; reject one since t cannot be negative)
when you have your value for t, sub in
in x-direction (origin at cannon, positive direction to right)
s = ut
range = 60cos45 x t
thus u have ur range.

nb: on ur diagram, it always helps to LABEL AXES (which way is positive?) and LABEL THE ORIGIN.
 

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