Question

A stone is thrown vertically upward with an speed of 40 m/s . The time interval for which particle was above 40 m from the ground is ( g = 10 m/s ^2)​

Answer 1

Given

stone is thrown vertically upward with an speed of 40 m/s .

To find

The time interval for which particle was above 40 m from the ground

Explanation

initial velocity ,u=40m/s

acceleration ,a=-g

Displacement,S=40m

Using the formula

$\boxed{\sf S= ut +\dfrac{1}{2}at^2}$

$= > 40 = 40t + ( - 10) \dfrac{1}{2} t ^{2}$

$= > 5 {t}^{2} - 40t + 40 = 0$

$= > {t}^{2} - 8t + 8 = 0$

Applying

$\boxed{\sf roots=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$t=\dfrac{-(-8)\pm \sqrt{(-8)^2-4(8)}}{2(1)}$

$t=\dfrac{8\pm \sqrt{64-32}}{2}$

$t =\dfrac{8\pm \sqrt{32}}{2}$

$t=\dfrac{8\pm 4\sqrt{2}}{2}$

$t=4\pm 2\sqrt{2}$

$\boxed{\sf t=4-3.4=0.6 s , t=4+3.4=7.4 s}$

So the body is at 40 m from the ground at t=0.6 s and 7.4 s

Some more formulas:

➡v²-u²=2as

➡v=u+at

➡S=ut+1/2at²