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

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

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

Since, Vy = 0, equation becomes

Substituting the values in the above equation

                                      S = 197.99 m

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

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: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg