Question

A helicopter is ascending vertically with a speed of 5.77 m/s . At a height of 102 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? [Hint: What is v0 for the package?] Express your answer to three significant figures and include the appropriate units.

Answer 1

ANS> The initial position of the package is 102 m above the earth surface , its final position is at 0 m.

The initial velocity of the package is 5.77 m/s since it was initially ascending with the helicopter,

the acceleration is -9.8m/s^2 which is the acceleration due to gravity.

Now we will use a basic kinematics equation to solve the above problem:

where, y(t) = height at any time t

y(0) = initial height

v(0) = initial speed

g = acceleration due to gravity

t = time

given:

y(t)= 0

y(0) = 102 m

v(0) = 5.77 m/s

g = -9.8 m/s^2

then substituting the values in the above equation we will take out the time

by rearranging we get a quadratic equation

by Sreedhar Acharya method ;

we get the values of t as 5.189 and -4.011.

Since time cannot be negetive then the value of t is 5.189 s.

The time  taken by the package to reach the ground is 5.189 s.