Solution of Assignment no . 7 - 1 Solution of Assignment no...

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1 Solution Initial condition: at t=0 Starting from the origin O (0, 0) * Equation of the path is given by: ( ) ) 1 .... ( tan 1 v 2 gx tan x y 2 2 o 2 α α + = Since points A & B lies on the path then they satisfy equation (1), substituting the x & y coordinates of both we get: A ( x A = 4, y A = 1) ( ) ( ) ( ) ( ) ) 2 .... ( 0 1 tan 4 tan 1 v 80 tan 1 v 2 4 10 tan 4 1 tan 1 v 2 gx tan x y 2 2 o 2 2 o 2 2 2 o 2 A A A = + + + = + = α α α α α α B ( x B = 6, y B = 0) ( ) ( ) ( ) ( ) ( ) ) 3 .... ( 30 tan v tan 1 0 tan 6 tan 1 v 180 tan 1 v 2 6 10 tan 6 0 tan 1 v 2 gx tan x y 2 o 2 2 2 o 2 2 o 2 2 2 o 2 B B B α α α α α α α α = + = + + = + = Substitute equation (3) in equation (2), we get: ) 4 .... ( 37 4 3 tan 0 1 tan 4 tan 30 80 o = = + α α α α 7 . Assignment no Solution of Free projectile motion 1. A projectile is fired from point O and passes through the points A (4, 1) and B (6, 0). Determine the firing velocity 0 v , the firing angle α , the time required to reach A and B, and velocities at A and B. α B(6,0) x A(4,1) y O
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2 A(4,1) s m / 3248 . 6 s m / 5764 . 1 A θ s m y B / 7364 . 4 . = s m x B / 3248 . 6 . = B(6,0) B θ Also Substitute equation (4) in equation (3), we get: ) 5 .... ( / 906 . 7 5 . 62 4 3 1 30 4 3 2 2 2 s m v v v o o o = + = * Time to reach point A: ( ) sec 632 . 0 5 4 906 . 7 4 cos = = = A A o A t t v x α * Time to reach point B: ( ) sec 948 . 0 5 4 906 . 7 6 cos = = = B B o B t t v x α * Velocity at point A: ( ) o A A A A A A A A o A o A o o A x y s m y x v s m gt v gt y y s m v x x 14 tan / 518 . 6 / 5764 . 1 632 . 0 10 5 3 906 . 7 sin / 3248 . 6 5 4 906 . 7 cos . . 2 . 2 . . . . . = = + = = = = = = = = = θ θ α α * Velocity at point B: ( ) o B B B B B B B B o B o B o o B x y s m y x v s m gt v gt y y s m v x x 37 tan / 902 . 7 / 7364 . 4 948 . 0 10 5 3 906 . 7 sin / 3248 . 6 5 4 906 . 7 cos . . 2 . 2 . . . . . = = + = = = = = = = = = θ θ α α
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3 2. The highest altitude of a projectile is (17 mt) and its range is (17 mt ) . Determine the firing velocity 0 v , the firing angle α , the time of flight, and the alternative firing angle to attain the same range. Solution We know that the range of a projectile is: ) 1 .... ( 170 cos sin 2 cos sin 2 2 2 = = α α α α o o v g v R Also we know that the maximum height of a projectile is: ) 2 .... ( 340 sin 2 sin 2 2 2 2 max = = α α o o v g v y Dividing equation (2) by equation (1), we get: ) 3 .... ( 76 4 tan 170 340 cos sin 2 sin 2 2 2 o o o v v = = α α α α α Substitute equation (3) in equation (2), we get: ) 4 .... ( / 19 16 340 17 340 17 4 2 2 2 s m v v v o o o × = = α O v o A B v A v B Θ Θ B Θ A x y 0 υ O y x 17m 17m
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4 3. A ball is projected from point A with a velocity s / m 15 0 = υ , which is perpendicular to the incline shown.
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