Question

a particle is projected with speed √20m/s at what time it will be at height 1.2m from ground
(g=10m/s^2

Answer 1

Answer:

$\displaystyle{t=0.054 \ sec \ or \ t=4.418 \ sec.}$

Explanation:

Given :

Initial velocity ( u ) = √ 20 m / sec.

Height ( h ) = 1.2 m

Acceleration ( g ) = 10 $m/sec^2$

We have to find time ( t )

We have  

$\displaystyle{s=ut-\frac{1}{2} gt^2}$

put the value here we get

$\displaystyle{1.2=\sqrt{20}\times t-\frac{1}{2}\times10\times t^2}\\\\\\\displaystyle{1.2=\sqrt{20}\times t-5\times t^2}\\\\\\\displaystyle{5 \ t^2-\sqrt{20} \ t+1.2=0}\\\\\\\displaystyle \text{On solving this we get }\\\\\\\displaystyle{t=0.054 \ sec \ or \ t=4.418 \ sec.}$

Thus we get answer.