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h_g

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a package is to be dropped from a plane flying at a height of 300m to a base on ground. If the plane is flying at 75ms-1, calculate the distance before the base, that the package must be released in order to reach the base.
 

photastic

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Ux = 75m/s
Uy = 0m/s
Consider downwards positive
ay = 9.8m/s/s , Sy = 300
Sy = Uyt + ½ayt^2
300 = 0 + 4.9t^2
t = square root (300/4.9)

Sx = Uxt
Sx = 75 x square root (300/4.9)
Therefore distance = 587m

Edit: Fizzy :)
 
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Fizzy_Cyst

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Ux = 75m/s
Uy = 0m/s
Consider downwards positive
ay = 9.8m/s/s , Sy = 300
Sy = Uyt + ½ayt
300 = 0 + 4.9t
t = 300/4.9

Sx = Uxt
Sx = 75 x 300/4.9
Therefore distance = 4592m
you forgot to square the t in your equation!

Sx = 75 x sqrt(300/4.9)

Sx = 586.85m
 

Fizzy_Cyst

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What he did above but replace

Sy = Uyt + ½ayt

with

Sy = Uyt + ½ayt^2

then t = sqrt (300/4.9) then follow my answer
 
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