## A ball thrown vertically upward with an initial speed of 2.8 from a window that is 3.6m above the ground. How long after the first ball is thrown should a second ball be simply dropped from the same window so that both balls hit the ground at the same?

Mar 6, 2017

${t}_{2} = 0.332$[s]

#### Explanation:

The first ball describes the path given by

${y}_{1} = {h}_{0} + {v}_{1} {t}_{1} - \frac{1}{2} g {t}_{1}^{2}$

At time ${t}_{2}$ the second ball is released so

${y}_{2} = {h}_{0} - \frac{1}{2} g {\left({t}_{2} - {t}_{1}\right)}^{2}$

and both hit the ground at the same time.

$\left\{\begin{matrix}{h}_{0} + {v}_{1} {t}_{1} - \frac{1}{2} g {t}_{1}^{2} = 0 \\ {h}_{0} - \frac{1}{2} g {\left({t}_{2} - {t}_{1}\right)}^{2} = 0\end{matrix}\right.$

Solving for ${t}_{1} , {t}_{2}$ we obtain a set of four solutions. The feasible one is

t_1 = (v + sqrt[2 g h + v^2])/g, t_2 = (-sqrt[2] sqrt[g^3 h] + g (v + sqrt[2 g h + v^2]))/g^2 giving

${t}_{1} = 1.188$ when both balls hit the ground and ${t}_{2} = 0.332$ the time when the second ball is released.

Attached a plot showing both paths.

Mar 8, 2017

$0.33 s$, rounded to two decimal place

#### Explanation:

Time taken by the second ball to hit ground after it is just dropped is governed by the kinematic equation
$h = u t + \frac{1}{2} g {t}^{2}$ ........(1)
Inserting given values and taking $g = 9.8 m {s}^{-} 2$, we get
$3.6 = 0 \times t + \frac{1}{2} \times 9.8 {t}_{2}^{2}$
$\implies {t}_{2}^{2} = \frac{3.6}{4.9}$
Ignoring the negative root as time can not be negative we get
$\implies {t}_{2} = \frac{6}{7} = 0.857 s$

Time taken by the first ball to hit ground after it is thrown vertically up is also governed by the kinematic equation (1). However, in this case gravity is acting against the direction of motion and distance is also negative as ball hits the ground. We get
$- 3.6 = 2.8 {t}_{1} + \frac{1}{2} \times \left(- 9.8\right) {t}_{1}^{2}$
$\implies 49 {t}_{1}^{2} - 28 {t}_{1} - 36 = 0$
This quadratic equation can be solved using the formula
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

I used inbuilt graphics tool to find out the roots. Valid root for time is

${t}_{1} = 1.189 s$
If both balls are to hit ground together, second ball should be released after time
${t}_{1} - {t}_{2}$
$= 1.189 - 0.857$
$= 0.33 s$, rounded to two decimal place