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PHYSICS HELP PLEASE VELOCITY BALL?
Could someone please explain to me how to do this
1) A ball is shot from the top of a building with an initial velocity of 20 m/s at an angle θ = 40° above the horizontal.
(a) What are the x and y components of the initial velocity?
vx = ____ m/s
vy = _____ m/s
(b) If a nearby building is the same height and 55 m away, how far below the top of the building will the ball strike the nearby building?
______ m
2 Answers
- JulianLv 69 years agoFavorite Answer
1
vx = 20cos40° = 15.32 m/s
vy = 20sin40° = 12.86 m/s
2
x = 55 = 20tcos40 ==> t =3.59s
y = 20tsin40 - gt²/2 = - 17 m
answer = 17 m below the top of the nearby building
- reddiceLv 44 years ago
you comprehend the thank you to define the placement of the middle of mass, splendid? x_cm = [ (mA)(xA) + (mB)(xB) ] / [ mA + mB ] and because velocity is the time by-made of the placement, v_cm = dx/dt = [ (mA1)(dxAdt) + (mB)(dxB/dt) ] / [ mA + mB ] so v_cm = [ (mA)(vA) + (mB)(vB) ] / [ mA + mB ] v_cm = [ (8.0 kg)(+5.6 m/s) + (2.5 kg)(-6.8 m/s) ] / [ 8.0 kg + 2.5 kg ] = +2.6 m/s to the two considerable figures that are justified. i think you do "only multiply then mass and velocity of each ball seperatley" (and then divide by using the completed mass), yet now you notice why.
