Physics question- projectile motion (1 Viewer)

[elfanger]

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]

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

 

[speersy]

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 ::+]

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.

 

[elfanger]

Thanks guys, i ended up getting it. Cool, now it seems really easy lol