MHB Work energy principle and power

[Shah 72]

A football is kicked from the ground level with speed 15m/s and rises to a height of 1.45m. Assume that air resistance is negligible
a) Find the speed of the ball when it is 1m above the ground
Increase in GPE= loss of KE
mgh=1/2mu^2-1/2mv2
10×(1.45-1)= 1/2×15^2-v^2/2
V= 14.7 m/ s
But textbook ans is 14.3m/s

Show that the ball was kicked at an angle of 21.0 degree
10× 1.45 sin theta= 1/2× 15^2- 1/2× 14.7^2
I don't get the ans.

 

[I like Serena]

The speeds and heights are mixed up.
Initially we have $v_0=15$ and $h_0=0$.
At $h_1=1\,\text{m}$, we have the unknown $v_1$ that we want to find.
And at $h_2=1.45\,\text{m}$, we have another unknown $v_2$, although we do know that the vertical speed is $v_{y,2}=0$ at this point.

Consider that the horizontal speed $v_x$ is constant.
In this case we can actually apply conservation of energy separately for both the horizontal and the vertical direction.
That is, at all times we have $\text{Total energy} =mhg + \frac 12 mv^2 = mhg + \frac 12 m(v_x^2 + v_y^2) = mhg + \frac 12 m v_x^2 + \frac 12 m v_y^2$, where $v_x$ is constant. Consequently the term with $v_x$ is the same each time, and we can remove it from the both sides of the corresponding conservation of energy equations.

Can we find the formulas for the total initial energy, the total energy at $h_1=1\,\text{m}$, and the total energy at $h_2=1.45\,\text{m}$, and fill in what we can?

 

[Shah 72]

The speeds and heights are mixed up.
Initially we have $v_0=15$ and $h_0=0$.
At $h_1=\,\text{m}1$, we have the unknown $v_1$ that we want to find.
And at $h_2=1.45\,\text{m}$, we have another unknown $v_2$, although we do know that the vertical speed is $v_{y,2}=0$ at this point.

Consider that the horizontal speed $v_x$ is constant.
In this case we can actually apply conservation of energy separately for both the horizontal and the vertical direction.
That is, at all times we have $\text{Total energy} =mhg + \frac 12 mv^2 = mhg + \frac 12 m(v_x^2 + v_y^2) = mhg + \frac 12 m v_x^2 + \frac 12 m v_y^2$, where $v_x$ is constant. Consequently the term with $v_x$ is the same each time, and we can remove it from the both sides of the corresponding conservation of energy equations.

Can we find the formulas for the total initial energy, the total energy at $h_1=1\,\text{m}$, and the total energy at $h_2=1.45\,\text{m}$, and fill in what we can?

So I get v= 14.7m/s

 

[I like Serena]

So I get v= 14.7m/s

That answer was deduced from $mgh_2 - mgh_1=\frac 12mv_0^2 - \frac 12 m v_1^2$, which is invalid, because $\frac 12mv_2^2$ is missing.

Instead we should have $mgh_0+\frac12 mv_0^2=mgh_1+\frac12 mv_1^2=mgh_2+\frac12 mv_2^2$, and we should also use that $v_x^2+v_y^2=v^2$ with constant $v_x$.

 

[Shah 72]

So according to what you said I get the ans 15.3m/s

 

[I like Serena]

So according to what you said I get the ans 15.3m/s

Can you show how you got there?

 

[Shah 72]

I u

That answer was deduced from $mgh_2 - mgh_1=\frac 12mv_0^2 - \frac 12 m v_1^2$, which is invalid, because $\frac 12mv_2^2$ is missing.

Instead we should have $mgh_0+\frac12 mv_0^2=mgh_1+\frac12 mv_1^2=mgh_2+\frac12 mv_2^2$, and we should also use that $v_x^2+v_y^2=v^2$ with constant $v_x$.

I understood. Thanks a lott!

 

[Shah 72]

Can you show how you got there?

Initial GPE= mgh= 0J
Initial KE= 225/2J
Middle GPE= mgh= 10J
Middle KE= 1/2 mv^2
Final GPE= 14.5J
Final KE = 0J
So I calculated for Initial and Middle one
0+ 225/2=10+v^2/2
So I get v=14.3m /s when h=1m

 

[Shah 72]

Can you show how you got there?

Can you also pls pls guide me with the next question where I have to prove that the ball is kicked at an angle of 21.0 degrees with the horizontal.

 

[I like Serena]

Final KE = 0J

The final KE is not 0. The velocity still has the constant horizontal speed component $v_x$.
So the final KE is $\frac 12mv_x^2$.

 

[I like Serena]

Can you also pls pls guide me with the next question where I have to prove that the ball is kicked at an angle of 21.0 degrees with the horizontal.

If we calculate the constant $v_x$, we can find the initial angle.

 

[Shah 72]

If we calculate the constant $v_x$, we can find the initial angle.

So you mean I have to calculate the horizontal distance?
Can you pls help me

 

[Shah 72]

The final KE is not 0. The velocity still has the constant horizontal speed component $v_x$.
So the final KE is $\frac 12mv_x^2$.

Let me try to work it out

 

[Shah 72]

If we calculate the constant $v_x$, we can find the initial angle.

So final velocity = 13.6m/s
How do I find the angle?

 

[I like Serena]

So final velocity = 13.6m/s
How do I find the angle?

The initial horizontal speed is the same as the final speed.
The cosine of the initial angle is the horizontal speed divided by the total initial speed.

 

[Shah 72]

The initial horizontal speed is the same as the final speed.
The cosine of the initial angle is the horizontal speed divided by the total initial speed.

Iam finding it a bit difficult. Can you pls show me a little bit of working

 

[I like Serena]

Let $\theta$ be the initial angle.
Then according to the definition of the cosine, we have
$$\cos\theta=\frac{v_x}{v_0}$$
The initial speed is given as $v_0=15$ and you have just calculated the final speed, which is horizontal, and the same as the initial horizontal speed $v_x$.

 

[Shah 72]

Let $\theta$ be the initial angle.
Then according to the definition of the cosine, we have
$$\cos\theta=\frac{v_x}{v_0}$$
The initial speed is given as $v_0=15$ and you have just calculated the final speed, which is horizontal, and the same as the initial horizontal speed $v_x$.

Thank you so so so so so so so much!