Force, mass and acceleration

In summary, the archer must shoot the arrow horizontally at the 40*m target, but it drops 2.2m over the range. The helicopter lifts the vehicle of mass 1500*kg at a vertical acceleration of 3.0*m*s–2.f
  • #1
1)One archery competition requires archers to fire a total of 90 arrows for a maximum possible score of 900. The targets are 1.22 m in diameter at 60, 50 and 40 metres distance. You can model the motion of the arrow to find out what problems the archer faces.

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?

[ I've 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. I've 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. I am 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.

[completely 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)
  • #2
1. y(position)=y0+(v0)*t-1/2*g*t^2


y0 is the heights of the arrow before its shot
v0 is the velocity of the arrow before its shot
and t is the time at which you wish to know the y(position)

2. Add gravity to the acceleration of of the copter.

3. Its the same as 2, but at an angle.
  • #3
thank you so much mate your a life saver however can i ask one more thing to you or others viewing. Why is it that you add the gravity to the acceleration of the hele? if the heles going up gravitys working on it to push it down.

Your 100% right and got the right answer so I am not questioning you but my theory is clearly wrong, can i just wana know where I've gone wrong.

Thank again mate, helped me alot.

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