Ball launched from a height of 5 meters --- Projectile motion

[Thermofox]

Homework Statement

A rubber ball, initially at a height $h = 5 m$ from the ground, is thrown horizontally with a velocity equal to $v_0 = 10 m/s$, as shown in the figure.
1) Being $O$ the origin of the frame of reference determine the distance $l_0$, namely where the ball first touches the ground.
2) Assuming that the impact with the ground is not perfectly elastic, but that both components of the ball's velocity are reduced by 20% in the impact, calculate the maximum height $h_1$ reached by the ball after the first bounce.
3) Under the assumptions of the previous question, calculate the distance $l_1$.

Relevant Equations

$x(t)= x_0 + vt$
$y(t)= y_0 + v_{0,y} + \frac 1 2 a_y t^2$
$\Delta E_m = 0$

I conceptually know how to solve this problem, what I struggle with is the direction of the acceleration.
For example to solve the first question I need to find the horizontal displacement when the ball hits the ground.
Therefore $l_0= x(t_1)= x_0 + v_0 t_1$, where $t_1$ is the moment the ball hits the ground. Thus to determine $l_0$ I first need to determine $t_1$. That it's not a problem since I can use $y(t)= y_0 + v_{0,y} + \frac 1 2 a_y t^2$.
In particular $y(t_1)=0 \Rightarrow 0 = h + 0\space t_1 +\frac 1 2 a_y t_1^2$. The only acceleration that the ball experiences is $g$, but $g$ is directed downwards $\Rightarrow a_y= -g$. Which means that $-h= -\frac 1 2 gt_1^2$. From this I can derive that:$$t_1= \sqrt{\frac{2h} {g}}$$

If I chose as the positive direction of $a_y$ the downward one, then I would have that $a_y=g$ and then $t_1= \sqrt{\frac{-2h} {g}}$ which doesn't make sense, so where is the error?

 

[Steve4Physics]

...since I can use $y(t)= y_0 + v_{0,y} + \frac 1 2 a_y t^2$.

You have a mistake in the above equation. But it doesn't affect your subsequent working.

If I chose as the positive direction of $a_y$ the downward one, then I would have that $a_y=g$ and then $t_1= \sqrt{\frac{-2h} {g}}$ which doesn't make sense, so where is the error?

With the 'downwards is positive' convention and the origin a $O$, what would be the value of $h$ in your above equation?

 

[kuruman]

The only acceleration that the ball experiences is $g$, but $g$ is directed downwards $\Rightarrow a_y= -g$. Which means that $-h= -\frac 1 2 gt_1^2$.

No, it does not mean that. You have the general kinematic equation
$y(t)= y_0 + v_{0,y}t + \frac 1 2 a_y t^2$
If you substitute $a_y=-g$, and for $v_{0,y}=0$, you get $~y(t)= y_0 - \frac 1 2 g t^2.$ This equation says that as time increases, the initial height $y_0$ decreases until it becomes zero.

 

[Thermofox]

You have a mistake in the above equation. But it doesn't affect your subsequent working.

oops! A $t$ jumped away. It should be $y(t)= y_0 + v_{0,y} t + \frac 1 2 a_y t^2$

With the 'downwards is positive' convention and the origin a $O$ , what would $h$ be the value of in your above equation?

$h$ would then be negative, I didn't consider that the positive direction is determined by the y-axis.

To do the second point I need to find $h_1$. For this I need to use $h_1=y(t_1)= y_0+v_{y,after impact} - \frac 1 2 gt^2$. Here the acceleration goes up, so it should be $... + \frac 1 2 gt^2$ right?

 

[Thermofox]

No, it does not mean that. You have the general kinematic equation
$y(t)= y_0 + v_{0,y}t + \frac 1 2 a_y t^2$
If you substitute $a_y=-g$, and for $v_{0,y}=0$, you get $~y(t)= y_0 - \frac 1 2 g t^2.$ This equation says that as time increases, the initial height $y_0$ decreases until it becomes zero.

I wrote $-h= -\frac 1 2 gt_1^2$ because $y(t)=0$ and $y_0=h$

 

[kuruman]

Here the acceleration goes up, so it should be $... + \frac 1 2 gt^2$ right?

Wrong. Is an object moving up not attracted by the Earth? Is the acceleration not in the same direction as the net force? You are confusing velocity with acceleration.

 

[Thermofox]

You are confusing velocity with acceleration.

You just made me realize that, thanks.

 

[kuruman]

I wrote $-h= -\frac 1 2 gt_1^2$ because $y(t)=0$ and $y_0=h$

When you write $y(t)$ you mean "the height above ground at any time $t.$ The ball makes contact with the ground at time $t_c=\sqrt{\dfrac{2h}{g}}$. The height above ground is zero at the specific time $t_c$, so you write $y(t_c)=0.$

 

[Thermofox]

When you write $y(t)$ you mean "the height above ground at any time $t.$ The ball makes contact with the ground at time $t_c=\sqrt{\dfrac{2h}{g}}$. The height above ground is zero at the specific time $t_c$, so you write $y(t_c)=0.$

Yes, I just had taken it for granted. I should've written $y(t_1)$, because I had defined $t_1$ as how you defined $t_c$

 

[kuruman]

Note that this problem can be solved more efficiently by using energy considerations instead of messing with the kinematic equations.

• If the ball starts with zero vertical velocity at height $h$ above ground, what is its vertical velocity $v_y$ just before it hits the ground.
• Given that its vertical velocity is $0.8v_y$ when it leaves the ground after the collision, how high does it rise?

 

[Thermofox]

If the ball starts with zero vertical velocity at height $h$ above ground, what is its vertical velocity $v_y$ just before it hits the ground.

$E_i=E_f \Rightarrow mgh= \frac 1 2 mv_{y,\text{right before collision}}^2$. Therefore $v_{y,bc}= \sqrt{2gh}$.

Given that its vertical velocity is $0.8v_y$ when it leaves the ground after the collision, how high does it rise?

$E_{\text{right after collision}} = E_f \Rightarrow \frac 1 2 m(0.8v_y)^2 = mgh_1$ $\Rightarrow h_1= \frac {\frac 1 2 (0.8v_y)^2} g$

I never considered that you can do an energetic analysis per component of velocity. In fact to solve the second point I first found $v_{bc}$ by doing $mgh+\frac 1 2 mv_0^2= \frac 1 2 mv_{bc}^2$, then I found $v_{x,ac} = 0.8v_x$and I can find it because $v_x$ is constant and finally I found $v_{y,ac}= \sqrt{0.8v_{bc}^2 - 0.8v_{x}^2}$. Successively I found how much time it takes to reach $h_1$ and only then I was able to find $h_1$.

It is definitely a faster approach. Unfortunately I think that I still have to use kinematics to solve point 3.
I solved it by first finding how much time, $t_3$, it takes for the ball to reach $h_1$. Thus $l_1= l_0 + v_{x,ac} (t_{flight})$, where $t_{flight}= 2t_3$. Also finding $t_3$ once you already know $h_1$ It's also faster.

 

[kuruman]

Even quicker
$mgh=\frac{1}{2}mv_{y,bc}^2\implies v_{y,bc}^2=2gh.$
$mgh_1=\frac{1}{2}mv_{y,ac}^2=\frac{1}{2}m(0.8v_{y,bc})^2=\frac{1}{2}m\times0.64\times2gh=0.64mgh\implies h_1=0.64h.$

For part 3 you already know that $t_1=\sqrt{\dfrac{2h}{g}}.$ The second bounce has a time of flight
$t_2=2\times\sqrt{\dfrac{2h_1}{g}}$ which is twice the time it takes to fall from rest from height $h_1$. Knowing the horizontal velocities for the first half-bounce and for the second full bounce, you can easily find the total horizontal distance.

 

[Thermofox]

Even quicker
$mgh=\frac{1}{2}mv_{y,bc}^2\implies v_{y,bc}^2=2gh.$
$mgh_1=\frac{1}{2}mv_{y,ac}^2=\frac{1}{2}m(0.8v_{y,bc})^2=\frac{1}{2}m\times0.64\times2gh=0.64mgh\implies h_1=0.64h.$

For part 3 you already know that $t_1=\sqrt{\dfrac{2h}{g}}.$ The second bounce has a time of flight
$t_2=2\times\sqrt{\dfrac{2h_1}{g}}$ which is twice the time it takes to fall from rest from height $h_1$. Knowing the horizontal velocities for the first half-bounce and for the second full bounce, you can easily find the total horizontal distance.

Wonderful, thanks again!