Posted by DJ on .
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. The plane is traveling horizontally with a speed of 280 km/h (77.8 m/s).
(b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m 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?
(c) With what speed do the supplies land in the latter case?
The question does not include part (a). I hope this is intended.
The problem can be summarized as follows:
Horizontal distance, H=425 m
Vertical distance, V=-235 m
horizontal velocity, h = 77.78 m/s
time to reach target, t = H/h = 5.464 s
The vertical component of velocity, v, is such that when the projectile is ejected upwards with a velocity v, it should reach V=-235 in 5.464 seconds.
The kinematics equation gives us:
V = vt + (1/2)(-g)t²
Solving for v give v=-16.2 m/s
The negative sign means the supplies should be given a kick downwards.
Check my calculations.
A recent solution to a similar problem is at your disposal :