# Question #d3dcb

Jun 12, 2015

It takes the ball $1.41 s$ to return to its thrower's hands.

#### Explanation:

For this problem, we will consider that no friction is involved

Let us consider the height from which the ball was launched as $z = 0 m$

The only force applied to the ball is its own weight:

$W = m \cdot g \leftrightarrow F = m \cdot a$

therefore, if we consider $z$ rising when the ball gets higher, the ball's acceleration will be

$- g = - 9.81 m \cdot {s}^{- 2}$

Knowing that $a = \frac{\mathrm{dv}}{\mathrm{dt}}$ then

$v \left(t\right) = \int a \cdot \mathrm{dt} = \int \left(- 9.81\right) \mathrm{dt} = - 9.81 t + c s t$

The constant value is found with $t = 0$. In other words, $c s t$ is the speed of the ball at the beginning of the problem. Therefore, $c s t = 6.9 m \cdot {s}^{- 1}$

$\rightarrow v \left(t\right) = - 9.81 t + 6.9$

Now, knowing that $v = \frac{\mathrm{dz}}{\mathrm{dt}}$ then

$z \left(t\right) = \int v \cdot \mathrm{dt} = \int \left(- 9.81 t + 6.9\right) \mathrm{dt}$
$= - \frac{9.81}{2} {t}^{2} + 6.9 t + c s t$

This time, $c s t$ is the ball's height at the beginning of the problem, assumed to be 0m.

$\rightarrow z \left(t\right) = - \frac{9.81}{2} {t}^{2} + 6.9 t$

Now, we want to find the time it takes the ball to rise to its maximum height, stop, then fall back to its starting height. We do that by resolving the following equation:

$- \frac{9.81}{2} {t}^{2} + 6.9 t = \left(- \frac{9.81}{2} t + 6.9\right) t = 0$

One obvious answer is $t = 0$ but it is pointless to specify that the ball starts from its starting point.

$- \frac{9.81}{2} t + 6.9 = 0$
$\rightarrow \frac{9.81}{2} t = 6.9$
$\rightarrow t = \frac{6.9 \cdot 2}{9.81} = \frac{13.8}{9.81} \approx 1.41 s$