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 H = 245 m below. If the plane is traveling horizontally with a speed of 257 km/hr (71.4 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 = 409 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? With what speed do the supplies land in the latter case?

Explanation / Answer

You have already solved the first part of the problem so I am solving the second part of the problem.

Now we solve for time using the question x = vox(t).

Range = x = 409 m and vox = initial horizontal velocity = 71.4 m/s.

therefore -
409 = 71.4 * (t)
=> t = 409 / 71.4 = 5.73 s

Now, using this time, we can solve for voy using the equation y = yo + (voy)t - (0.5)gt^2.

We know:
y = 0
yo = 245 m
voy = ?
t = 5.73 s
g = 9.8 ms^2

put the values -
0 = 245 + voy(5.73) - (0.5)(9.8)(5.73)^2
=> voy(5.73) = -84.12

=> voy = -14.7 m/s

Second Part of the problem -

We know that the supplies' final horizontal velocity is 71.4 m/s because horizontal velocity is always constant. However, we need to solve for final vertical velocity (vy):

vy = voy - gt
vy = (-14.7) - (9.8)(5.73)
=> vy = - 70.85 m/s

Now you use the x and y components of final velocity to find the object's final speed:

R = sqrt(vy^2 + vx^2)
R = sqrt( (-70.85)^2 + (71.4)^2 )
R = 100.6 m/s

So, the final speed at which the supplies drops = 100.6 m/s

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