Hi, I need help solving a problem from the Giancoli Textbook:Chapter 3, Problem

Question

Hi, I need help solving a problem from the Giancoli Textbook:Chapter 3, Problem 35. However, you need the actual textbook tolook at the figure this problem refers to. Please help. Thankyou!

Figures on Page 67:

35. A rescue plane wants to drop supplies to isolated mountainclimbers on a rocky ridge 235m below. If the plane is travelinghorizontally with a speed of 250km/h (69.4 m/s):

a) How far in advance of the recipients (horizontal distance)must the goods be dropped (Figure 3-37a)?

b) Suppose, instead, that the plane releases the supplies ahorizontal distance of 425m in advance of the mountain climbers.What vertical velocity (up or down) should the supplies be given sothat they arrive precisely at the climber’s position (Figure3-37b)?

c) With what speed o the supplies land in the latter case?

Explanation / Answer

Okay. Lets say the speed of the plane being 69.4m/s, isVx. the vertical distance to the climber is Y. acceleration is A. time is T. the initial vertitical velocityis Vyi and the distance dropped ahead of time is D. We need to solve for time in part A first. using the equationY=(Vyi)T+(1/2)A(T^2) We see T is equal to ((2Y/a)) WE multiply T by Vx tocalculate D. In part be the designate a set D. Dividing this by Vx givesus T. plug T back into the initial equation we used in part Aand solve for Vyi. To solve part C we must calculate the final velocity in the Ydirection (Vyf). Vyf= (Vyi)^2+2A(Y) After solving for this we now have 2 components of final velocityVyf and Vx. To calculate final speed: Final speed = (Vyf^2 + Vx^2)

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