I'm not sure what it's called. here is the question.
A ball is thrown 50 m/s at 15 degrees above horizontal,then 30 deg,45 deg,60 deg, and 75 deg avove horizontal. What vertical distance(height) and horizontal distance does the ball travel each time?
yeah! we were doing this at the beginning... and it is very confusing! i admit!
*drags out Physics notes* lol
ummm... ok,
well you know you have to make a right triangle with the verticle and horizontal parts, and i dont know how i'm going to explain this without drawing it to show you, but i can try!
anyways... first you have to find the measurent of the horizontal (H) component which you use cosign for!
i'll do the 15 degrees one for you
so you get...
cos15=
H.......... 50 m/s
H= 50 m/s x cos15
H= 48.3
and now for verticle (V) use sin
sin15=
V..........50 m/s
V= 50 m/s x cos15
V= 12.9
so for 15 degrees you get a total horizontal distance of 48.3m and a total vertical distance of 12.9m
i hope that helped some!!!!! i'm horrible with explaining things to people! just do the 30, 45, 60 and 75 degree angles the same way! just substitute the 15's with whatever angle you're doing!
tell me if you have any more questions!