Question

A helicopter is ascending at a rate of 12m/s. At a height of 80m above the ground, a package is dropped. how long does the package take to reach the ground?​​

Answer 1

Answer:

5.45 second it takes to reach ground

Answer 2

When package is dropped from helicopter, it has the same velocity as that of the helicopter. So,

Initial velocity of package (u) = 12 m/s

Height above the ground from where package is dropped (h) = 80 m

Acceleration due to gravity (g) = 10 m/s²

By using second equation of motion, we get:

$\bf \leadsto h = ut + \dfrac{1}{2} g {t}^{2} \\ \\ \rm \leadsto - 80 = 12t + \dfrac{1}{2} \times( - 10) {t}^{2} \\ \\ \rm \leadsto - 80 = 12t - 5 {t}^{2} \\ \\ \rm \leadsto - 16 = 2.4t - {t}^{2} \\ \\ \rm \leadsto {t}^{2} - 2.4t - 16 = 0 \\ \\ \rm \leadsto t = \dfrac{2.4 \pm \sqrt{ 5.76 + 64 } }{2} \\ \\ \rm \leadsto t = \dfrac{2.4 \pm \sqrt{ 69.76} }{2} \\ \\ \rm \leadsto t = \dfrac{2.4 + 8.4 }{2} \\ \\ \rm \leadsto t = \dfrac{10.8}{2} \\ \\ \rm \leadsto t = 5.4 \: s$

$\therefore$ Time taken by package to reach the ground (t) = 5.4 s