Question

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4m/s), (a) how far in advance of the recipients ( horizontal distance) must the goods be dropped. (b) Suppose, instead that the plane release the supplies a horizontal distance of 425m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climber's position? (c) with what speed do the supplies land in the latter case

Answer 1

Answer:

Explanation:

For vertical displacement of food-supply

if time taken to reach the ground be t

235 = 1/2 g t²  , initial velocity downwards is zero.

t = 6.92 s

Horizontal displacement of food during this period

= 6.92 x 69.4 m

= 480.25 m .

b )

time required to cover horizontal distance of 425 m

= 425 / 69.4 = 6.124 s  

This time period will be the time of vertical fall of 235 m . Let initial vertical velocity required be u

h = ut + 1/2 gt²

235 = u x 6.124 + .5 x 9.8 x 6.124²

235 = u x 6.124 + 183.76

u = 8.36 m /s

c )

v = u + gt

= 8.36 + 9.8 x 6.124

= 68.37 m /s

This will be vertical component of velocity .

horizontal velocity = 69.4 m /s

resultant velocity = √ ( 68.37² + 69.4²)

= √(4674.45 +4816.36)

= 97.42 m /s