Question

A helicopter is ascending vertically at a speed of 19.6 ms'. When it is at a height of 156.8 m above the ground, a stone is dropped. How long does the stone take to reach the ground? ​

Answer 1

$initial velocity :$ $Vi = -19.6 ms-1$

$height : s = 156.8 m$

gravitional accerlation  = 9.8 ms-²

$\time t = ?$

Using second equation of motion ::  s = vi t + 1/2 at²

                                                          156.8 = (-19.6×t) + 1/2 ×(9.8)×t²

                                                            156.8 = -19.6t  + 4.9 t²

$Dividing both side by 4.9$

156.8/4.9 = -19.6t/4.9 + 4.9 t²/4.9

 32 =  - 4t - 32  = 0

 t² - 4t - 32  = 0

 t²  - 8t +4t -32 = 0

t (t-8) + 4 (t-8) = 0

(t+4)  (t - 8) = 0

(t+4)  = 0  ⇒ t = - 4

(t -8) = 0  ⇒ t = +8

Time is always  positive  , so  the time taken by  the stone  to reach  the ground  will be  $t = 8s$

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