Question

A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? (hint: the package initial velocity equals the helicopter velocity)

Answer 1

Here it is given that initial speed of the package will be same as speed of the helicopter

$v_i = 5.10 m/s$

displacement of the package as it is dropped on ground

$d = -105 m$

acceleration is due to gravity

$a = -9.8 m/s^2$

now by kinematics

$y = v* t + \frac{1}{2}at^2$

$-105 = 5.1 * t - \frac{1}{2}*9.8*t^2$

$4.9 t^2 - 5.1 t - 105 = 0$

by solving above equation we have

$t = 5.2 s$

so it will take 5.2 s to reach the ground

Answer 2

The time required by the package to reach the ground will be 5.12 seconds.

Given:

The helicopter  is ascending vertically with a speed of $5.10\rm \; m/s$.

Now, the package is dropped from the height of $h=-105\rm \; m$. Height is negative because it is decreasing.

Initial velocity of the package is $u=5.1\rm \; m/s$.

The acceleration acting on the package will be, $a=-g=-9.8\rm \; m/s^2$. acceleration is negative because it is acting downwards.

Use the second equation of motion to find the time required to reach the ground.

$h=ut+\dfrac{1}{2}at^2\\-105=5.1t-\dfrac{1}{2}\times 9.8\times t^2\\4.9t^2-5.1t-105=0$

Solve the above quadratic equation to find the value of time as,

$4.9t^2-5.1t-105=0\\t=\dfrac{5.1\pm \sqrt{5.1^2+4\times 4.9\times 105}}{2\times 4.9}\\t=5.12\rm \; s$

Therefore, the time required by the package to reach the ground will be 5.12 seconds.

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