# A stone is thrown vertically up from the tower of height 25m with a speed of 20m/s time taken to reach the ground?

Oct 23, 2017

The stone will reach the ground in 5 seconds after being thrown upwards from the tower.

#### Explanation:

When the stone is thrown vertically upwards,

$u$ = initial velocity =$20$m/s

It will reach some height where its final velocity will become zero,

$v = 0$m/s

$a = - g = - 10 m {s}^{- 2}$ ----negative as it is going in direction against the gravitational pull. (Consider $g = 10 \frac{m}{s} ^ 2$ for ease of calculation).

Height = distance it travels upwards will be :

$s = \frac{{v}^{2} - {u}^{2}}{2 a}$

$s = \frac{0 - {20}^{2}}{- 2 \times 10}$ m

$s = - \frac{400}{-} 20$m

$s = 20$m

Time taken to reach this distance will be :

$t = \frac{v - u}{a}$

$t = \frac{0 - 20}{-} 10 = 2$

$t = 2$ seconds ----------let this be ${t}_{1}$

So it reaches $20$m from the peak of tower of height $25$m. So to reach to ground,the stone has to cover:

$s = 20 + 25 = 45$m.

When it starts falling from $45$m above ground, its initial velocity is zero, so:

$u = 0$

We know it will come down with uniform acceleration

$a = g = 10 m {s}^{-} 2$

We know : $s = u t + \frac{1}{2} a {t}^{2}$

$45 = 0 \times t + \frac{1}{2} \times 10 \times {t}^{2}$

$45 \times \frac{2}{10} = {t}^{2}$

${\cancel{45}}^{9} \times \frac{2}{\cancel{10}} ^ 2 = {t}^{2}$

$9 \times \frac{\cancel{2}}{\cancel{2}} = {t}^{2}$

$\therefore {t}^{2} = 9$, i.e. $t = 3$ seconds-----Let this be ${t}_{2}$

Total time taken after throwing the stone upwards will be :

$t = {t}_{1} + {t}_{2}$

$t = 2 + 3 = 5$seconds

$\therefore$The stone will reach the ground in 5 seconds after being thrown upwards from the tower.