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The arrow leaves the bow at 60 m*s–1 and travels at almost this speed horizontally for the whole of its flight.

The arrow, of course, falls because of the acceleration due to gravity. You can find its position at any moment by working out how far it has moved horizontally and how far it has fallen vertically.

1. The archer shoots the arrow horizontally at the 40*m target, How far does it drop over this range?

[ Ive worked out it takes 0.66 seconds to reach the target. So gravity should on arrow should be 9.8/3 x 2 = 6.5M. Ive clearly gone wrong somewhere, because the answer is 2.2m. unfortanly AS OCR B course resources at utter rubbish.]

2)The helicopter lifts the vehicle of mass 1500*kg at a vertical acceleration of 3.0*m*s–2. Show that the tension in the cable is 1.9x10^4

(theres a diagram showing the length of the cable to be 50m if this is any help)

[ I assumed it would just be f = m x a = 1500 x3 = 4500. im abit lost on it here, is it something to do with vectors and if so how can you work out 2 vectors going in oppisite directions with one being a mass and not a magnitude]

At a later time, the helicopter is moving forward in level flight at a constant velocity of 50 m s–1. The cable carrying the vehicle now hangs at a steady angle of 15°

Calculate the tension in the cable in this case.

[completly lost, the problem is that the as OCR b book is mixed up, nothing on free fall and gravity is mentioned in the whole book but questions on the CD appear everywhere)