Question

A helicopter is ascending vertically upward with a speed of 10 m/s. At a height 12 m above the earth,a package is dropped from the window. How much time does it take for the package to reach the ground ?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Time\:taken=2.843\:sec}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Height = 12 \: m \\ \\ \tt: \implies Initial \: speed = - 10 \: m/s \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Time \: taken \: to \: reach \: package \: on \: ground =?$

• According to given question :

$\tt \circ \: Acceleration = 10 \: {m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies 12 = - 10 \times t + \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \tt: \implies 12 = - 10t + 5 {t}^{2} \\ \\ \tt: \implies {5t}^{2} - 10t - 12 = 0 \\ \\ \tt: \implies t = \frac{ - ( - 10) \pm \sqrt{ {( - 10)}^{2} - 4 \times 5 \times - 12} }{2 \times 5} \\ \\ \tt: \implies t = \frac{10 \pm \sqrt{100 + 240} }{10} \\ \\ \tt: \implies t = \frac{10 \pm \sqrt{340} }{10} \\ \\ \tt: \implies t = \frac{10 \pm18.43}{10} \\ \\ \tt: \implies t = \frac{10 + 18.43}{10} \\ \\ \green{\tt: \implies t = 2.843 \: sec} \\ \\ \green{\tt \therefore Time \: taken \: to \: reach \: ground \: is \: 2.843 \: sec}$