Solution of Assignment no . 7

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

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Unformatted text preview: 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 ....( 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 tan 6 tan 1 v 180 tan 1 v 2 6 10 tan 6 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 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 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 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 . 5 4 906 . 7 4 cos = = ⇒ = A A o A t t v x α * Time to reach point B: ( ) sec 948 . 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 . 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 . 10 5 3 906 . 7 sin / 3248 . 6 5 4 906 . 7 cos . . 2 . 2 . . . . . ≅ ⇒ = = + = − = − = − = − = = = = = θ θ α α 3 2. The highest altitude of a projectile is (17 mt) and its range is (17 mt ) . Determine the firing velocity 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...
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This note was uploaded on 12/21/2010 for the course ENGINEERIN mp108 taught by Professor Elbarki during the Spring '08 term at Alexandria University.

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

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