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3 questions about projectile
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(1) In vertical direction, initial velocity = 75.2 sin 34.5 = 42.6 m/s Applying the formula s = ut + at2/2 with u = 42.6, a = -9.8 and t = 1.5, we have s = 52.9 s With s = 0, i.e. vertical displacement = 0, we can find out the time for returning to the ground which is: ut + at2/2 = 0 u + at/2 = 0 42.6 - 4.9t = 0 t = 8.69 s Horizontal distance travelled = 8.69 x 75.2 cos 34.5 = 539 m (2) Referring to the diagram below: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Sep08/Crazyvector6.jpg Applying the sine law: sin β = 100 sin 125/600 β = 7.85 Thus, θ = 180 - 125 - β = 47.15 Finally, the plane should head at a direction og 90 - 47.15 = 42.85 north of east to achieve the required path. (3) The relative velocity of the car to the train is 20 km/h forward, thus to run 1 km more than the train for the car, time required = 1/20 hours or 3 mins. If they are in opp. directions, relative velocity = 170 km/h and time required = 1/170 hours or 0.35 mins.
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3 questions about projectile
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1. A projectile is fired with an initial speed of 75.2m/s at an angle of 34.5degree above the horizontal on a long flat firing range. What is the velocity of the projectile 1.5s after firing.* Maximum height reached by the projectile=92.6m the total time in the air= 8.69s the total horizontal... 顯示更多 1. A projectile is fired with an initial speed of 75.2m/s at an angle of 34.5degree above the horizontal on a long flat firing range. What is the velocity of the projectile 1.5s after firing. * Maximum height reached by the projectile=92.6m the total time in the air= 8.69s the total horizontal distance covered= 539m 2.An airplane, whose air speed is 600km/h, is supposed to fly in a straight path 35.0 degree north of east. But a steady 100km/h wind is blowing from the north. In what direction should thw plane head? 3.An automobile travelling 95km/h overtakes a 1.00-km-long train travelling in the same direction on a track parallel to the road. If the train's speed is 75km/h, how long does it take the car to pass it and how far will the car have traveled in this time?What are the results if the car and train are travelling in opposite directions? 更新: well,,,,i still don't know understand the first one.. can u explain more ??? by the way...your explainations are very clear in the other two最佳解答:
(1) In vertical direction, initial velocity = 75.2 sin 34.5 = 42.6 m/s Applying the formula s = ut + at2/2 with u = 42.6, a = -9.8 and t = 1.5, we have s = 52.9 s With s = 0, i.e. vertical displacement = 0, we can find out the time for returning to the ground which is: ut + at2/2 = 0 u + at/2 = 0 42.6 - 4.9t = 0 t = 8.69 s Horizontal distance travelled = 8.69 x 75.2 cos 34.5 = 539 m (2) Referring to the diagram below: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Sep08/Crazyvector6.jpg Applying the sine law: sin β = 100 sin 125/600 β = 7.85 Thus, θ = 180 - 125 - β = 47.15 Finally, the plane should head at a direction og 90 - 47.15 = 42.85 north of east to achieve the required path. (3) The relative velocity of the car to the train is 20 km/h forward, thus to run 1 km more than the train for the car, time required = 1/20 hours or 3 mins. If they are in opp. directions, relative velocity = 170 km/h and time required = 1/170 hours or 0.35 mins.
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