lecture6 - Let's write these two equations in terms of vo...

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Let’s write these two equations in terms of v o and θ : 2 2 1 ) (sin t a t v y y y o o f + + = θ t v x o ) (cos θ = Δ If we can find the time using the top equation, then we can plug it into the bottom equation to get Δ x . 2 2 1 ) 8 . 9 ( ) 50 )(sin 00 . 9 ( 19 . 1 75 . 0 t t o - + = 0 44 . 0 89 . 6 9 . 4 2 = - - t t Solve using quadratic formula, where a = 4.9, b = -6.89, and c = -0.44: ) 9 . 4 ( 2 ) 44 . 0 )( 9 . 4 ( 4 ) 89 . 6 ( 89 . 6 2 - - - ± = t 8 . 9 49 . 7 89 . 6 ± = s 0.06 - or s 47 . 1 = t Now that we have t , we can solve for Δ x : t v x o ) (cos θ = Δ s) 47 . 1 )( 50 cos m/s)( 00 . 9 ( o = m 50 . 8 = Let’s see how well we did!
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Example : As an airplane takes off, it climbs with a constant speed of 75 m/s. The plane’s velocity vector makes an angle of 15 deg. with the horizontal. Unfortunately, when the plane reaches an altitude of 2500 m during this climb, one of its engines falls off. How long does it take the engine to hit the ground below? v o = 75 m/s 15 o x y y o = 2500 m v oy y f = 0 m a y = -9.8 m/s 2 2 2 1 t a v y y y o o f y + + = m/s 4 . 19 ) 15 )(sin 75 ( 15 sin = = = o o o o v v y 2 9 . 4 4 . 19 2500 0 t t - + = ) 9 . 4 ( 2 ) 2500 )( 9 . 4 ( 4 ) 4 . 19 ( 4 . 19 2 - - - ± - = t ) 9 . 4 ( 2 222 4 . 19 - ± - = t s 6 . 24 or s, 7 . 20 = - = t t
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Example : A golf ball rolls off a horizontal cliff with an initial speed of 11.4 m/s. The ball falls a vertical distance of 15.5 m into a lake below. (a) How much time does the ball spend in the air? (b) What is the speed v of the ball just before it strikes the water? v o = 11.4 m/s y o = 15.5m 2 2 1 (a) t a t v y y y o o f y + + = 2 2 1 ) 8 . 9 ( 0 5 . 15 0 t - + + = 2 9 . 4 5 . 15 t - = - s 78 . 1 = t y f = 0 y = 0 y v v x = v ox v y = v fy 2 2 ) ( ) ( (b) fy ox v v v + = m/s 4 . 11 = ox v t a v v t v v a y oy fy oy fy y + = - = m/s 4 . 17 ) 78 . 1 )( 8 . 9 ( 0 - = - + = fy v m/s 8 . 20 ) 4 . 17 ( ) 4 . 11 ( 2 2 = - + = v
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3.4 Relative Velocity Let’s say that you are standing on the ground at rest beside a road, when a pickup truck drives by you at 25 m/s. Relative to you , the truck is moving at 25 m/s (Velocity of the truck relative to the ground) v TG = 25 m/s Now suppose that someone riding in the back of the truck can throw a baseball at 25 m/s. First consider the case when he throws the baseball forward, the same direction as the truck is moving. Now how fast is the baseball moving relative to you ??? v TG = 25 m/s v BT = 25 m/s TG BT BG v v v + = m/s 50 25 25 = + = The ball whizzes by you at 50 m/s!
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Now suppose that the person riding in the back of the truck throws the baseball at 25 m/s off the back of the truck i.e. in the opposite direction that the truck is moving. Now how fast is the baseball moving relative to you ???
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