Question : I need help with this assignment?
An airplane traveling 1013 m above the ocean at 100 km/h is to drop a box of supplies to shipwrecked victims below.(a) How many seconds before being directly overhead should the box be dropped?(b) What is the horizontal distance between the plane and the victims when the box is dropped?6.Divers in Acapulco dive from a cliff that is 60 m high. If the rocks below the cliff extend outward for 25 m, what is the minimum horizontal velocity a diver must have to clear the rocks? 7.A dart player throws a dart horizontally at a speed of 12.3 m/s. The dart hits the board 0.32 m below the height from which it was thrown. How far away is the player from the board? 8.A student is playing with a radio-controlled race car on the balcony of a sixth-floor apartment. An accidental turn sends the car through the railing and over the edge of the balcony. Does the time it takes the car to fall depend upon the speed it had when it left the balcony?
- asked by ahsanvip1@sbcglobal.net
All Answers:
Answer #1
you need to have a slurpee
- answered by chimchim
Answer #2
#8 If the car was going fast surely it would "fly" out further then if it just dropped off so maybe it could be a few seconds more if it was going faster, a way to test this would be to simply time a marble rolling off a desk as a slow speed and then time it at pace! I am not into physics so I dont really know just an idea! Good Luck!
- answered by diamonds10002001
Answer #3
#8. No. It's hard to believe but gravity exerts the same force on an object travelling at high speed as it does on one travelling at low speed. (If the car flies upwards off of the porch the velocity would make a difference. It wouldn't if it flies straight off.)
- answered by no_nonsense
Answer #4
(a)Dy = 1/2A*t^2 + Vy*t, but Vy = 0, so t = sqrt(Dy/(1/2 A))Dy = -1013 m, A = -g (both negative since downward)t = sqrt(1013/4.9) = 14.38 s(b) Dx = Vx*t Vx = 100 km/h = 27.78 m/sDx = 27.78*14.38 = 399.4 m6. Dy = 1/2A*t^2 + Vy*tA = -g, Dy = -60 m, Vy is not specified. If they jump up when they dive, they have more time to clear the rocks. But we have to assume they just run straight off the edge, so Vy = 0, and t = sqrt(60 m/(1/2*g)) = 3.5 s.To go > 25 m, Dx = Vx*t > 25 m, so Vx > 25 m/3.5 s = 7.14 m/s7. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (thrown horizontally),so t = sqrt(Dy/(1/2*g)) = sqrt(0.32/4.9) = 0.256 s Dx = Vx*t = 12.3 m/s*0.256 s = 3.14 m8. Dy = 1/2 A*t^2 + Vy*t.By now you should see that the motion in the y (vertical) direction depends only on the initial velocity in the y direction, the distance to fall in the y direction, and the acceleration in the y direction. The speed in the x direction has nothing to do with it.
- answered by almond9090
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