# Ball is thrown at the edge of the roof of a building

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1. Jan 15, 2017

### rosekaty

1. The problem statement, all variables and given/known data

A ball is thrown at the edge of the roof of a building in a direction at an "alpha" angle above the horizontal. It landed 5 seconds later at 50 meters from the building. The maximum height reached by the ball during its trajectory is 20 meters above the roof.

Find The initial velocity and the angle with which it was launched .
2. Relevant equations

3. The attempt at a solution
I found the equations:
y(x)=-0.5(g*t^2+Vo*sin(alpha)*t + L (with L the height of the building(that I don't have btw))
but I can't go anywhere without L or alpha.

2. Jan 15, 2017

### Staff: Mentor

Look at the horizontal and vertical components of the motion separately to begin with. Can you determine the components of the initial velocity from the information given?

3. Jan 15, 2017

### PeroK

You don't actually need any numbers to solve a problem. So, if a quantity is missing, you must solve the problem algebraically until you find an equation for that quantity in terms of quantities for which you do have the numbers.

Hint: you will probably have to generate two simultaneous equations here, One equation using the known maximum height above the launch point (20m) and one equation using the range (50m).

4. Jan 15, 2017

### rosekaty

Yes, I thought about that and I got:

the launch point = x= -L
or x=Vo^2*((2*L*cos^2*alpha+sin(2*alpha)))/g

the maximum height =(vo^2*sin^2(alpha))/2*g

5. Jan 15, 2017

### rosekaty

Because of the total height of the building that I don't have, I don't think I can.
If I "cut" the motion to the top of the building to a point on the motion in the same horizontal, Delta x still missing even if I have Delta y which is the 20meters.

6. Jan 15, 2017

### Staff: Mentor

That's not a valid argument. The horizontal and vertical motions for the trajectory are entirely independent. For example, the ball travels 50 m horizontally. How long did it take? So what's the horizontal speed?

7. Jan 15, 2017

### rosekaty

The horizontal speed is 10 m*s^-1 then.

8. Jan 15, 2017

### zexxa

Another tip you may consider is the Pythagoras theorem, it might come in handy for you.
You know that the velocity can be resolved into 2 components, one in the y-axis and the other in the x-axis, what's more they relate to each other with this equation. $v^2=a^2+b^2$ where $a$ is the vertical component and $b$ the horizontal.

As @gneill mentioned above, you can determine the horizontal speed. I see you've already found the way to determine the vertical speed in the post above his as well. Just pair them up and viola there you have it!

9. Jan 15, 2017

### Staff: Mentor

Correct!

Now consider the vertical motion alone. What information are you given that pertains to it?

10. Jan 15, 2017

### PeroK

A "viola" is a musical instrument! I think "voila" is the word you want.

11. Jan 15, 2017

### zexxa

Ay, you're right my bad LOL Wrote it out of habit haha

12. Jan 15, 2017

### rosekaty

Well I think the maximum height can be useful. Maybe if I use this equation: the maximum height =(vo^2*sin^2(alpha))/2*g

13. Jan 15, 2017

### Staff: Mentor

That is the range equation and pertains to horizontal distance.

If you're looking at vertical motion only there will be no angle involved, just the initial vertical velocity and the height achieved.

EDIT: My mistake. That's not the range equation (although it looked like it to me when I looked at it all too quickly!). It will in fact give you the maximum height, but employs the initial launch velocity and launch angle, neither of which you have yet. See my later post below for more information.

Last edited: Jan 15, 2017
14. Jan 15, 2017

### rosekaty

Vertical velocity= -g*t+ maximum height ?

or (y(x)-L-(0.5*g*t^2))/t=Vertical velocity

(To be honest, I think I'm lost. I see your point but there is so many equations I don't know which one I should use.)

15. Jan 15, 2017

### zexxa

No worries
That equation of yours is wrong however. Velocity is measured in $m/s$. The units for $gt$ is $m/s$ but the units for your height achieved is in $m$. The dimensions are wrong here.
I suppose you meant vertical velocity=initial vertical velocity + $gt$ but this equation isn't especially useful for you now because you do not know the time taken for it to reach its peak, where you know that it's vertical velocity at that point is 0.

There is another equation that you can use for this case $v^2=u^2+2as$ where $v$ is the final velocity, $u$ the initial velocity, $a$ the acceleration and $s$ the displacement it underwent. You know the vertical velocity at the top of the throw (i.e. 0), the vertical displacement it underwent, and the acceleration it had so you can determine the initial velocity in the vertical component.

16. Jan 15, 2017

### Staff: Mentor

Oops. I've given you some incorrect information in my last post. That equation in ascii formatting reminded me of the range equation and I jumped to an incorrect conclusion. That formula will give you the height, starting with the initial velocity and angle of launch.

The thing is, in this instance you will be looking at some initial vertical velocity alone with no angle. Your equation extracts the vertical component from the initial velocity using $v_o sin(\alpha)$ and then squares it. You don't need to do that. Just use $v_y$ as the vertical velocity and square that.

17. Jan 15, 2017

### rosekaty

Okay so:
final velocity^2=Initial velocity^2+2*a*H
0=Initial velocity+2*(-9.8)*20
Initial velocity = 19.79 m*s^-1

18. Jan 15, 2017

### zexxa

Yes that is right, but is that the answer you're looking for?

19. Jan 15, 2017

### rosekaty

No, so if I use what you told me earlier
Vo=22.18 m*s^-1

20. Jan 15, 2017

### zexxa

Yep that's it!
Then for the next part of the question to find the angle $\alpha$, how do you plan to use the information you have at hand to solve it?