posted by Kim .
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge H = 231 m below. If the plane is traveling horizontally with a speed of 227 km/h (63.1 m/s), how far in advance of the recipients (horizontal distance) must the goods be dropped?
Suppose, instead, that the plane releases the supplies a horizontal distance of x = 325 m in advance of the mountain climbers. What vertical velocity (use the positive direction as upwards) should the supplies be given so that they arrive precisely at the climbers' position? And at what speed do the supplies land ?
To fall 231 m, the time required is
t = sqrt(2H/g) = 6.87 seconds.
The supplies will travel forward
6.87 * 63.1 = 433.2 m
Your answer is correct.
For the second part, the time to fall must be changed to (325 m)/V = 5.15 seconds
The supplies must then be given an initial (negative) velocity v downwards so that it reaches the ground in less time. Solve this equation
-231 = -4.9 t^2 +v*t
= -130.0 + 5.15 v
v = -101/5.15 = -19.7 m/s
The supplies land at velocity v - g*t
where t = 5.15 s
Aerodynamic friction has been neglected.
It's going to hit the ground very fast.