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Published byMatteo Biby Modified over 2 years ago

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Over the Edge Horizontal Projectiles

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A red ball rolls off the edge of a table What does its path look like as it falls? Parabolic path

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As the red ball rolls off the edge, a green ball is dropped from rest from the same height at the same time Which one will hit the ground first? They will hit at the SAME TIME!!!

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The same time?!? How?!? v ix The red ball has an initial HORIZONTAL velocity (v ix ) But does not have any initial VERTICAL velocity (v iy = 0) The green ball falls from rest and has no initial velocity IN EITHER DIRECTION! v iy and v ix = 0

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One Dimension at a Time Both balls begin with no VERTICAL VELOCITY Both fall the same DISTANCE VERTICALLY Find time of flight by solving in the appropriate dimension We can find an object’s displacement in EITHER DIMENSION using TIME

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Example #1 A bullet is fired horizontally from a gun that is 1.7 meters above the ground with a velocity of 55 meters per second. At the same time that the bullet is fired, the shooter drops an identical bullet from the same height. –Which bullet hits the ground first? Both hit the ground at the same time

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Example #2 An airplane making a supply drop to troops behind enemy lines is flying with a speed of 300 meters per second at an altitude of 300 meters. –How far from the drop zone should the aircraft drop the supplies? Need time from vertical d y = v iy t + ½ a y t 2 300 m = 0 + ½ (-9.81 m/s 2 )t 2 t = 7.82 s Use time in horizontal d x = v ix t + ½ a x t 2 d x = (300 m/s)(7.82 s) + 0 d x = 2346 m

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Example #3 A stuntman jumps off the edge of a 45 meter tall building to an air mattress that has been placed on the street below at 15 meters from the edge of the building. –What minimum initial velocity does he need in order to make it onto the air mattress? Need time from vertical d y = v iy t + ½ a y t 2 45 m = 0 + ½ (-9.81 m/s 2 )t 2 t = 3.03 s Use time to find v v = d / t v = 15 m / 3.03 s v = 4.95 m /s

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Example #4 A CSI detective investigating an accident scene finds a car that has flown off the edge of a cliff. The car is 79 meters from the edge of the 25 meter high cliff. –What was the car’s initial horizontal velocity as it went off the edge? Need time from vertical d y = v iy t + ½ a y t 2 25 m = 0 + ½ (-9.81 m/s 2 )t 2 t = 2.26 s Use time in horizontal d x = v ix t + ½ a x t 2 79 m = v ix (2.26 s) + 0 v ix = 34.96 m/s

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