標題:
physics---projectile
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?
最佳解答:
Vertical motion and horizontal motion are independent.Firstly, consider the horizontal motion, the projectile travels at constant horizontal speed. Horizontal speed, vh = 75.2cos34.5* = 61.974 ms-1 Lastly, we consider the vertical motion, take upward direction be positive. Initial speed, u = 75.2sin34.5 ms-1 Time of flight, t = 1.5 s Acceleration due to gravity, g = -10 ms-2 By v = u + gt v = 75.2sin34.5* + (-10)(1.5) vv = 27.594 ms-1 Therefore, the resultant speed = √(vh2 + vv2) = √[61.9742 + 27.5942] = 67.8 ms-1 Let x be the angle between the velocity and the horizontal tanx = vv / vh = 27.594 / 61.974 x = 24.0* So, the velocity of the projectile 1.5s after firing is 67.8 ms-1 in the direction 24.0* above the horizon.
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physics---projectile
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發問: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?
最佳解答:
Vertical motion and horizontal motion are independent.Firstly, consider the horizontal motion, the projectile travels at constant horizontal speed. Horizontal speed, vh = 75.2cos34.5* = 61.974 ms-1 Lastly, we consider the vertical motion, take upward direction be positive. Initial speed, u = 75.2sin34.5 ms-1 Time of flight, t = 1.5 s Acceleration due to gravity, g = -10 ms-2 By v = u + gt v = 75.2sin34.5* + (-10)(1.5) vv = 27.594 ms-1 Therefore, the resultant speed = √(vh2 + vv2) = √[61.9742 + 27.5942] = 67.8 ms-1 Let x be the angle between the velocity and the horizontal tanx = vv / vh = 27.594 / 61.974 x = 24.0* So, the velocity of the projectile 1.5s after firing is 67.8 ms-1 in the direction 24.0* above the horizon.
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