Physics, 05.11.2019 05:31 brenda2512

Apackage falls out of a helicopter that is traveling horizontally at 70 m/s it falls into the water below 8.0 seconds later assuming no air resistance what is the horizontal distance it travels while falling

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Answer from: nana54muller

The horizontal distance the package traveled while falling, S = 197.99 m

Explanation:

Given,

The horizontal velocity of the helicopter, v = 70 m/s

The time taken by the package to fall into the water is, t = 8.0 s

The initial horizontal component of velocity, Vx = 70 m/s

The initial vertical component of velocity, Vy = 0 m/s

The time of flight is given by the formula

$t=\frac{Vy+\sqrt{Vy^{2} +2gh}}{g}$

Since Vy = 0 ; solving for h

h = gt²/2 meter

Substituting the given values in the above equation

h = (9.8 x 8²) / 2

= 39.2 m

The horizontal distance traveled while falling is given by the formula

$S=\frac{Vx[Vy+\sqrt{Vy^{2} +2gh}] }{g}$

Since, Vy = 0, equation becomes

$S=\frac{Vx\sqrt{2gh} }{g}$

Substituting the values in the above equation

$S=\frac{70\sqrt{2X9.8X39.2} }{9.8}$

S = 197.99 m

Hence, the horizontal distance the package travels while falling is, S = 197.99 m

Answer from: montgomerykarloxc24x

D = 560m

Explanation: The helicopter is traveling with a velocity of 70m/s in the x-axis, and we know that it takes 8 seconds to fall into the water.

Whem the helicopter "throws" the package, the only equations for the package are the initial velocity, that is the same of the helicopter (70 m/s in the x axis)

and the gravity acceleration in the y axis, but we only care for the horizontal distance, so this does not matter.

Then, knowing that the package travesl for 8 seconds with a velocity of 70m/s in the horizontal direction, the total distance traveled is:

70m/s*8s = 560m

Answer from: Quest

Ithink your answer is d

Answer from: Quest

the rearrangement of particles in a physical change. hope this

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Apackage falls out of a helicopter that is traveling horizontally at 70 m/s it falls into the water...

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