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

[rosekaty]

Homework Statement

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 .

Homework Equations

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.

 

[gneill]

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?

 

[PeroK]

Homework Statement

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 .

Homework Equations

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.

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).

 

[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

 

[rosekaty]

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?

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.

 

[gneill]

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.

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?

 

[rosekaty]

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?

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

 

[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!

 

[gneill]

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

Correct!

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

 

[PeroK]

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!

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

 

[zexxa]

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

 

[rosekaty]

Correct!

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

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

 

[gneill]

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.

 

[rosekaty]

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.

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.)

 

[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.

 

[gneill]

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.

 

[rosekaty]

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.

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

 

[zexxa]

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

 

[rosekaty]

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

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

 

[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?

 

[rosekaty]

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?

There is a lot of equations involivng the angle, I could pick the one with the maximum height that says:
H= (initial velocity^2*sin(alpha)^2)/2*g

Or tan(alpha) = Vo*sin(alpha)/Vo*cos(alpha)

 

[zexxa]

Not bad! Both methods work perfectly well and you're on the right track. Guess this problem is solved then :)

 

[rosekaty]

Not bad! Both methods work perfectly well and you're on the right track. Guess this problem is solved then :)

Well...I tried it and for the first equation I realize that I don't have the maximal height since I don't have the height of the building. And the second one seems..unsolvable.

 

[zexxa]

The height you should use in the first equation is not the absolute maximum height with respect to the ground, it is with respect to the release, which would be 20m.

For the second equation, what's the issue? I suppose you're inputting $v_0$ and solving for alpha on the tangent, sine and cosine equations, which is a bit tedious. You should realize first that $v_0sin\alpha$ is actually $v_y$ (which you calculated) and $v_0cos\alpha$ is $v_x$ (which you also calculated). Now you have just a simple equation to solve $tan\alpha$=$v_y$/$v_x$

 

[rosekaty]

The height you should use in the first equation is not the absolute maximum height with respect to the ground, it is with respect to the release, which would be 20m.

For the second equation, what's the issue? I suppose you're inputting $v_0$ and solving for alpha on the tangent, sine and cosine equations, which is a bit tedious. You should realize first that $v_0sin\alpha$ is actually $v_y$ (which you calculated) and $v_0cos\alpha$ is $v_x$ (which you also calculated). Now you have just a simple equation to solve $tan\alpha$=$v_y$/$v_x$

Yes thank you I just get that ! It was simple.

I found an angle of 1°

 

[zexxa]

Hahaha are you sure? Check your work again (or rather the way you presented your answer)

Hint: 1-something is correct if you look at it in a different way

 

[rosekaty]

Hahaha are you sure? Check your work again (or rather the way you presented your answer)

Hint: 1-something is correct if you look at it in a different way

Lol Yes I checked many times I'm pretty sure i's false. I did arctangent(19.79/10) because we found Vy=19.44 and Vx=10 m*s^-1

 

[rosekaty]

I GET IT ! I was in radian !

 

[zexxa]

Yep! That's the reason why you got a 1+ for your result.
Now that you know it was in radians, you can easily convert it into degrees and you're done! :)

 

[rosekaty]

Thank you all for your help ! Really I mean it, it wasn't easy for me and you helped me till the end. Double difficulty I'm French

 

[rosekaty]

Yes I got 63°

 

[zexxa]

Good job! And keep up the good work :)