A rescue plane wants to drop supplies to isolated mountain climbers on a rocky r

Question

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 below
a) if the plane is traveling horizontally with a speed of 230 (63.9 ), how far in advance of the recipients (horizontal distance) must the goods be dropped (see the figure )?

b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425 in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position (see the figure ?

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

Explanation / Answer

a: Given: s=235m v=62.5m/s The key to solving this problem is knowing that the time it takes a free falling body (ffb) to reach the rocky ridge is equal to the time it takes the supplies to reach the rocky ridge. So, use the equation s=ut+1/2at^2 where s= the vertical distance from the plane to the rocky ridge, u is the initial velocity, a the acceleration of gravity, and t the time. In the case of an ffb, u=0. a=9.8m/s^2 s=235m Substitute given values in the formula: 235=0*t+1/2*9.8*t^2 235=4.9t^2 t^2=235/4.9 =47.95 t=6.93s Now solve for the horizontal distance using the basic formula: s=vt s in this case is the horizontal distance, v the horizontal speed of the plane, and t the time it takes the supplies to reach the rocky ridge. Substitute known values in the formula: s=62.5*6.93 =433m in advance of the recipients. b)So now the supplies will overshoot the recipients by 8 m (433-425) unless it is given a vertical initial speed downwards, to cut the time it takes the supplies to reach the rocky ridge. Note that giving the supplies a vertical initial speed upwards will delay the time it will take for the supplies to reach the rocky ridge, and would overshoot it even further. By how much will the pilot have to cut the time so as to make the supplies reach the same spot at the rocky ridge? The answer is by as much time as it takes the plane to travel 8m. So we use the basic formula again: s=vt s=8m v=62.5m/s Substitute given values: 8=62.5t t=8/62.5 =0.128s Thus now the total time it should take the supplies to reach the ground should be 6.93-0.128 or 6.80s. We again go back to our formula: s=ut+1/2at^2 s=235m t=6.80s a=9.8m/s^2 Substitute known values: 235=u*6.80+1/2*9.8*6.80^2 235=6.80u+226.6 6.80u=235-226.6 =8.4 u=8.4/6.80 =1.24m/s c) We use the formula: v^2-u^2=2as where v= the final speed, or the speed the supplies land u= the initial speed a=9.8m/s^2 s=235m Substitute known values: v^2-1.24^2=2*9.8*235 v^2=4606+1.54 =4607.5 v=67.9m/s

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