# Question #ffc23

May 29, 2016

$\frac{5}{4} s \mathmr{and} 7.8125 m$

#### Explanation:

Let the balls meet t sec after the second ball is thrown.
So the 1st ball will then complete (t+1) sec of its journey.

The imitial velocity of the 1st ball is zero .So it will fall a height $h = \frac{1}{2} \cdot g \cdot {\left(t + 1\right)}^{2}$

The 2nd ball's initial velocity being $30 \frac{m}{s}$ it should fall the same height during t sec
So $h = 30 \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

Inserting $g = 10 m {s}^{-} 2$ and equating two height we can write
$\frac{1}{2} \cdot 10 \cdot {\left(t + 1\right)}^{2} = 30 t + \frac{1}{2} \cdot 10 \cdot {t}^{2}$
$\implies 5 {\left(t + 1\right)}^{2} - 5 {t}^{2} = 30 t$
$\implies 5 \left(2 t + 1\right) = 30 t$
$\implies 2 t + 1 = 6 t$
$\implies t = \frac{1}{4} s$

So the ball will meet $\left(t + 1\right) = \frac{1}{4} + 1 = \frac{5}{4} s$ after the 1st ball starts.

During this time it will fall
$h = \frac{1}{2} \cdot 10 \cdot {\left(\frac{5}{4}\right)}^{2} = \frac{125}{16} m = 7.8125 m$