Projectile motion questiondon't understand about angles and vectors

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SUMMARY

The discussion centers on a projectile motion problem from Giancoli's 5th edition, specifically chapter 3, problem 38. A rescue plane traveling horizontally at 69.4 m/s must drop supplies to climbers located 235 m below, requiring a vertical velocity of 8.41 m/s for the supplies to reach them precisely. The participant initially attempted to solve the problem using horizontal distance and time calculations, ultimately realizing that the angle of release does not affect the required vertical velocity for this scenario.

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Homework Statement


Giancoli 5th edition chapter 3 problem 38

a rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235m below
the plane is traveling horizontally with a speed of 69.4m/s
b) supposes, instead that the plane releases the supplies a horizontal distance of 425m in advance of the moutain climbers. What vertical velocity should the supplies be given so that they arrive precisely at the climbers position?




The Attempt at a Solution



I attempted it several times using two different method
the second time was using

[tex]X = V_0 t[/tex]
t= 6.12s

so then
[tex]y =y_0 + V_0t + 1/2 a_y (t^2)[/tex]

answer was 8.41m/s


but then my question is that
for my first attempt at this question
I try to figure out the angle if they threw it upward or downward
but why does that angle not matter for the vertical or horizontal velocity?
 
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O DUDE
I reread the question and it said it only gave vertical velocity
thats why
okay no need for help now
thx
 

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