## . Emergency rations are to be dropped from a plane to some stranded hikers. The search and rescue plane is flying at an altitude of 1500 m a

Question

. Emergency rations are to be dropped from a plane to some stranded hikers. The search and rescue plane is flying at an altitude of 1500 m at 70 m/s. a. Determine where the ideal drop point would be (measured horizontally from the hikers). (Answer is 1.22 km I need help solving b) b. The pilot notices (too late!) that they have passed the ideal drop point; the supplies now need to be launched vertically in order to land near the hikers. Calculate the vertical launch velocity required, given that the plane is now 1 km away (horizontally) from the hikers.

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3 years 2021-08-19T08:13:47+00:00 2 Answers 5 views 0

1. Answer: 35 m/s will be the initial vertical launch velocity.

Explanation:

So this is where learning about quadratic equations becomes useful. First we have have to calculate the amount of time it takes the basket to reach the hikers which can be solved by:

1000 m / (70 m/s) = 14.28 s [We know the horizontal displacement and velocity, so calculating time is simple. Convert km to m to make life easier, and remember to handle vertical and horizontal velocity individually.]

The hard part is to find vertical velocity. To do so, set up your equation:

d=v•t+ 1/2 at^2

Vertical displacement we know is 1500 m, time is 14.28s and half acceleration is 4.9m/s^2. Your equation should look like this when the numbers are plugged in:

1500(m)=v•×14.28(s)+4.9(m/s^2)×14.28(s^2)

To convert this into a solvable equation, we will set the left side to 0 and arrange the equation, to look like this:

0=v•14.28(s)+4.9(m/s^2)×14.28(s^2)-1500(m)

Isolate v• and solve.

(Note: v• stands for initial vertical velocity if confused)

35 m/s down

Explanation:

The horizontal speed of the package is 70 m/s.  So the time needed to reach the hikers is:

1000 m / (70 m/s) = 14.28 s

Taking down to be positive, the initial velocity needed is:

Δy = v₀ t + ½ at²

1500 m = v₀ (14.28 s) + ½ (9.8 m/s²) (14.28 s)²

v₀ = 35 m/s

The package must be launched down with an initial velocity of 35 m/s.