Question

0.11 An object is thrown vertically upwards with a velocity 50 m/s. At what time it will be at height 120 m?
(g = 10 m/s)​

Answer 1

velocity of the object is 50 metre per second ,

distance = 120m

by third equation of motion ,

S = ut + 0.5 at^2

120 = 50t – 10t^2

=> t^2 – 5t + 12

now ,

a = 1 , b = –5 , C = 12

$t = \frac{ - b \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{ 2a} \\ put \: the \: values \: of \: a \: b \: and \: c \: </strong><strong>you</strong><strong> \: will \: get \: 2 \: answers \: from \: which \: + ve \: one \: will \: be \: correct$

Hope it will help uh..

THNKS!!