# An object is thrown vertically from a height of 7 m at  4 ms^-1. How long will it take for the object to hit the ground?

May 7, 2016

Assuming the object has initial upward velocity $\to t = 1.67$ $s$
Assuming the object has initial downward velocity $\to t = 0.85$ $s$

#### Explanation:

We have been given the following informations

The object is thrown *with velocity $u = 4$ $m {s}^{-} 1$
from a height of $h = 7$ $m$ and if it takes time t sec to reach the ground then to apply the equation of motion we may go as follows

Assuming the object has initial upward velocity

• $u = + 4$ $m {s}^{-} 1 \to \text{upward taken +ve}$
• $h = - 7$ $m \to \text{downward taken -ve}$
• Acceleration due to gravity $g = - 9.8$ $m {s}^{-} 2 \to \text{downward taken -ve}$

So the equation, $h = u t + \frac{1}{2} g {t}^{2}$ becomes
$\implies - 7 = 4 \times t - \frac{1}{2} \times 9.8 {t}^{2}$
$\implies 4.9 {t}^{2} - 4 t - 7 = 0$

$t = \frac{- \left(- 4\right) + \sqrt{{\left(- 4\right)}^{2} - 4 \times 4.9 \left(- 7\right)}}{2 \times 4.9} = 1.67 s$ [negative value of t neglected]

Assuming the object has initial downward velocity

• $u = + 3$ $m {s}^{-} 1 \to \text{downward taken +ve}$
• $h = + 7$ $m \to \text{downward taken +ve}$
• Acceleration due to gravity $g = + 9.8 \frac{m}{s} ^ 2 \to \text{downward taken +ve}$

So the equation,$h = u t + \frac{1}{2} g {t}^{2}$ becomes
$\implies 7 = 4 \times t + \frac{1}{2} \times 9.8 {t}^{2}$
$\implies 4.9 {t}^{2} + 4 t - 7 = 0$

$t = \frac{- 4 + \sqrt{{\left(4\right)}^{2} - 4 \times 4.9 \left(- 7\right)}}{2 \times 4.9} = 0.85 s$ [negative value of t neglected]