# Question 5a794

Aug 3, 2017

$\text{initial velocity} = 19.6$ $\text{m/s}$

$\text{final velocity} = - 19.6$ $\text{m/s}$

#### Explanation:

We're asked to find the initial and final velocities of an object, given that its maximum height is $19.6$ $\text{m}$.

We realize that when a particle is at its maximum height, its instantaneous $y$-velocity ${v}_{y}$ is zero.

With that being said, we can use the kinematics equation

ul((v_y)^2 = (v_(0y))^2 + 2a_y(y - y_0)

where

• ${v}_{y}$ is the $y$-velocity

• ${v}_{0 y}$ is the initial $y$-velocity (what we're trying to find)

• ${a}_{y}$ is the vertical acceleration, equal to $- g$, or $- 9.8$ ${\text{m/s}}^{2}$

• $y$ is the height (maximum height, in meters)

• ${y}_{0}$ is the initial height (ground level, $0$ 'm")

We know:

• ${v}_{y} = 0$

• v_(0y) = ?

• ${a}_{y} = - 9.8$ ${\text{m/s}}^{2}$

• $y = 19.6$ $\text{m}$

• ${y}_{0} = 0$ $\text{m}$

Plugging in known values:

${\left(0\right)}^{2} = {\left({v}_{0 y}\right)}^{2} + 2 \left(- 9.8 \textcolor{w h i t e}{l} \text{m/s"^2)(19.6color(white)(l)"m" - 0color(white)(l)"m}\right)$

$0 = {\left({v}_{0 y}\right)}^{2} - 384.16 \textcolor{w h i t e}{l} {\text{m"^2"/s}}^{2}$

v_(0y) = sqrt(384.16color(white)(l)"m"^2"/s"^2) = color(red)(ulbar(|stackrel(" ")(" "19.6color(white)(l)"m/s"" ")|)

This is the initial velocity of the ball, so now let's find the final velocity, which we can use the same equation for, only this time $y$ and ${y}_{0}$ are both $0$ (when it touches the ground again, its height is $0$):

${\left({v}_{y}\right)}^{2} = \left(19.6 \textcolor{w h i t e}{l} \text{m/s"^2) + 2(-9.81color(white)(l)"m/s"^2)(0color(white)(l)"m" - 0color(white)(l)"m}\right)$

Which leaves us with

${\left({v}_{y}\right)}^{2} = {\left(19.6 \textcolor{w h i t e}{l} \text{m/s}\right)}^{2}$

or

${v}_{y} = \sqrt{{\left(19.6 \textcolor{w h i t e}{l} \text{m/s}\right)}^{2}}$

Since the object is on its way down as it touches the ground, we take the negative solution of this:

v_y = color(blue)(ulbar(|stackrel(" ")(" "-19.6color(white)(l)"m/s"" ")|)

Any speculations as to why the final and initial velocities have the same magnitude?

Aug 3, 2017

Consider the diagram

We can solve this question using one of the kinematic equation

color(brown)(v^2=u^2+2ax

Where $v$ is the final and $u$ is the initial velocity.$a$ is the acceleration and $x$ is the distance travelled

Let's consider the first case

rArrcolor(orange)(v=0 $\left(\text{as the ball comes to rest at the maximum height}\right)$

rArrcolor(orange)(u=?

rArrcolor(orange)(a=-9.8 m/s $\left(\text{acceleration due to gravity}\right)$

rArrcolor(orange)(x=19.6m

Plug all the values in the equation

$\rightarrow {v}^{2} = {u}^{2} + 2 a x$

$\rightarrow {0}^{2} = {u}^{2} + 2 \left(- 9.8\right) \left(19.6\right)$

$\rightarrow 0 = {u}^{2} + \left(- 384.6\right)$

$\rightarrow - {u}^{2} = - 384.6$

$\rightarrow u = \sqrt{384.6}$

color (green)(rArr=19.61m/s#

So, the final velocity will be the opposite of the initial velocity $= - 19.61 \frac{m}{s}$

Hope this helps!