# An object is thrown vertically from a height of 15 m at  6 m/s. How long will it take for the object to hit the ground?

Nov 7, 2017

2.4 seconds

#### Explanation:

If ${t}_{1}$ is the time taken by object to reach same height then

t_1 = (2u)/g = (2 × 6 ms^-1) / (9.8 ms^-2) = 1.2 s

Velocity of object when it hits the ground is

$v = \sqrt{{u}^{2} + 2 a S}$

v = sqrt((6 ms^-1)^2 + (2 × 9.8 ms^-2 × 15m))

$v = 18.2 m {s}^{-} 1$

If ${t}_{2}$ is further time taken by object to reach the ground then

${t}_{2} = \frac{v - u}{g} = \frac{18.2 m {s}^{-} 1 - 6 m {s}^{-} 1}{9.8 m {s}^{-} 2} = 1.2 s$

Total time taken $= {t}_{1} + {t}_{2} = 1.2 s + 1.2 s = 2.4 s$

Nov 12, 2017

The time is about 2.5 s.

#### Explanation:

Compared to your first answer, this is an alternate method of solving.

I will choose to have up be the positive direction. Use the kinematic (suvat) formula
$s = u \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}$
where

• $s = - 15 m$ because it will hit the ground - below the point it was thrown from
• $u = 6 \frac{m}{s}$ positive because it was thrown up
• $a = - g = - 9.8 \frac{m}{s} ^ 2$.

Now, plugging in the data and solving for t
$- 15 m = 6 \frac{m}{s} \cdot t + \frac{1}{2} \cdot \left(- 9.8 \frac{m}{s} ^ 2\right) \cdot {t}^{2}$

Adjust so that all 3 are on one side and zero is on the other (preparing for use of the quadratic equation)

$\frac{1}{2} \cdot 9.8 \frac{m}{s} ^ 2 \cdot {t}^{2} - 6 \frac{m}{s} \cdot t - 15 m = 0$

$4.9 \frac{m}{s} ^ 2 \cdot {t}^{2} - 6 \frac{m}{s} \cdot t - 15 m = 0$

$t = \frac{6 \pm \sqrt{{6}^{2} - 4 \cdot 4.9 \cdot \left(- 15\right)}}{2 \cdot 4.9}$

$t = \frac{6 \pm \sqrt{36 + 294}}{9.8}$

$t = \frac{6 \pm \sqrt{330}}{9.8} = \frac{6 \pm 18.17}{9.8}$

The - part of the $\pm$ will yield a negative time, so that solution is invalid. I will continue using the + part of the $\pm$.

$t = \frac{24.17}{9.8} = 2.47 s$

I hope this helps,
Steve