Question

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 189 km/h (52.5 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 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?

Answer 1

Step 1:

FInd time taken by supplies to reach ground, Using 2nd kinematic equation in vertical direction:

h = V0y*t + (1/2)*a*t^2

h = -235 m, a = -g = -9.81 m/sec^2

V0y = 0, since plane is traveling horizontally, So

-235 = 0*t + (1/2)*(-9.81)*t^2

t = sqrt (2*235/9.81) = 6.92 sec

Step 2:

find range in projectile motion:

R = V0x*t

V0x = Speed of plane in horizontal direction = 52.5 m/sec

So,

R = 52.5*6.92 = 363.3 m

SO Supplies should be dropped 363.3 meter in advance

Part B.

Now when supplies are dropped 425 m in advance and they should reach to climbers, then

R = V0x*t1

t1 = time of flight for supplies = R/V0x = 425/52.5 = 8.095 sec

Now using 2nd kinematic equation in vertical direction:

h = V0y*t1 + (1/2)*a*t1^2

h = -235 m, a = -g = -9.81 m/sec^2

V0y = ?

-235 = V0y*8.095 + (1/2)*(-9.81)*8.095^2

V0y = [9.81*8.095^2/2 - 235]/8.095

V0y = 10.68 m/sec

Since V0y is positive, So Velocity should be upward