# An object is thrown vertically upward from a height of 2 m at 1 m/s. How long will it take for the object to hit the ground?

Jan 1, 2017

The initial velocity of the object in upward direction $u = 1 \frac{m}{s}$
( upward direction is taken positive )

So the downward acceleration due to gravity $g = - 9.8 \frac{m}{s} ^ 2$

The object is thrown from height $2 m$. If it reaches ground after$t s$ then the displacement of the object during $t s$ will be $h = - 2 m$
Hence applying equation of kinematics for under gravity motion we can write
$h = u \times t + \frac{1}{2} \times g \times {t}^{2}$

$\implies - 2 = 1 \times t - \frac{1}{2} \times 9.8 \times {t}^{2}$

$\implies 4.9 {t}^{2} - t - 2 = 0$

$\implies t = \frac{1 + \sqrt{{\left(- 1\right)}^{2} - 4 \times 4.9 \times \left(- 2\right)}}{2 \times 4.9} \approx 0.75 s$