1. A projectile is launched from a 300 m high cliff at an angle of 30? with an i
ID: 1899851 • Letter: 1
Question
1. A projectile is launched from a 300 m high cliff at an angle of 30? with an initial velocity of 150 m/s. (g=9.8m/s ). Begin by drawing a diagram and determine: a) The total time of the flight b) The horizontal velocity of the rocket on impact with the ground. c) The vertical velocity of the rocket on impact with the ground. d) The horizontal distance to the impact location from the edge of the cliff e) Magnitude of the velocity on impact f) Direction of the velocity on impact (angle with horizontal)Explanation / Answer
a) vertical component if the initial velocity: Vh = 150×sin(30) = 75 m/s now find the total time of the flight by using d = Vi×t + a×t²/2 where d = height = -300 m (negative because below initial height) Vi = initial vertical velocity = 75 m/s t = time = ? a = acceleration by gravity = -9.8 m/s² so -300 = 75×t + (-9.8)×t²/2 4.9×t² - 75×t - 300 = 0 this is a quadratic equation that you can solve for t vD = v( (-75)² - 4×4.9×(-300)) = 107.26 t1 = (75 + 107.26) / (2×4.9) = 18.60 s t2 = (75 - 107.26) / (2×4.9) = -3.29 s a negative time makes no sense here, so there is only 1 valid solution: t = 18.60 s < - - - - answer b) Vh = 150×cos(30) = 129.90 m/s < - - - - answer c) (Vf)² = (Vi)² + 2×a×d where Vf = final vertical velocity = ? Vi = initial vertical velocity = 75 m/s a = acceleration by gravity = -9.8 m/s² d = height = -300 m (negative because below initial height) so (Vf)² = 75² + 2×(-9.8)×(-300) (Vf)² = 11505 Vf = -107.26 m/s < - - - - answer (it is falling down, so you must take the negative square root) d) r = Vh×t r = 129.9×18.60 = 2416.14 m < - - - - answer e) Vr = v( (-107.26)² + 75² ) = 130.88 m/s < - - - - answer f) ? = arctan( -107.26 / 75 ) = -55.03° < - - - - answerRelated Questions
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