Need help for a projectile problem

[mad]

Hello..
There is a projectile problem that I can't seem to resolve.. I have tried it for more than 30min now and can't find out how to do it.. I got an exam in 2 days so would someone please help me and tell me how to do this exercice:

A basket-ball is launched at an angle of 45 degres (it is on the ground). The basket is at an horizontal distance of 4m and a vertical distance of 0,8m. What is the initial velocity (Vo) required to reach the basket?

.. sorry for my english this is translated from french.. here is what I tried to do but I know its not good..

http://img39.exs.cx/img39/4460/gdg.jpg

thank you

 

[Tide]

First, set [itex]y_0 = 0[/tex] because they gave the height of the hoop relative to the starting height.<br /> <br /> You should be able to combine your first two equations to give y as a function of x and when you solve that equation for $v_0$ you will find<br /> $$v_0 = \sqrt \frac {g x^2}{x-y}$$<br /> which actually works out to a nice round number in your problem![/itex]

 

[mad]

First, set [itex]y_0 = 0[/tex] because they gave the height of the hoop relative to the starting height.<br /> <br /> You should be able to combine your first two equations to give y as a function of x and when you solve that equation for $v_0$ you will find<br /> $$v_0 = \sqrt \frac {g x^2}{x-y}$$<br /> which actually works out to a nice round number in your problem![/itex]

$<br /> Thanks a lot.. but would you explain me how you got that formula and what is wrong with what I did so I can learn to not do it again? <br /> <br /> Thank you very much for your time$

 

[Tide]

Since the launch angle is 45 degrees
$$y = \frac {v_0 t}{\sqrt 2} - \frac {1}{2} g t^2$$
$$x = \frac {v_0 t}{\sqrt 2}$$
Solve the second equation for t and substitute into the first:
$$y = x - \frac {g x^2}{v_0^2}$$
from which my earlier equation follows.

 

[Tide]

P.s.

As to what you did wrong I think you just got bogged down in your algebra.

 

[mad]

Since the launch angle is 45 degrees
$$y = \frac {v_0 t}{\sqrt 2} - \frac {1}{2} g t^2$$
$$x = \frac {v_0 t}{\sqrt 2}$$
Solve the second equation for t and substitute into the first:
$$y = x - \frac {g x^2}{v_0^2}$$
from which my earlier equation follows.

I feel very dumb for asking this.. but where does $${\sqrt 2}$$ come from?

I know that X=Vxo*t which gives X=Vo * Cos45* t
and Y= Yo +Vyo*t - 1/2 *g * t^2

but why do you put a $${\sqrt 2}$$ ?

Thanks very much for your time

 

[faust9]

$$\cos 45^\circ=\sin 45^\circ=\frac{1}{\sqrt 2}=\frac{\sqrt 2}{2}$$

 

[mad]

YES! I finally got it! I was doing it alright but I just messed up with my algebra like you said.. except you gave me a much simpler way. I just couldn't use the $${\sqrt 2}$$ because my teacher never told use about that.

Also, when do I have to use this: 2 sin A * cos A = Sin 2A ? In what type of problems is this used?
thanks again

 

[Tide]

YES! I finally got it! I was doing it alright but I just messed up with my algebra like you said.. except you gave me a much simpler way. I just couldn't use the $${\sqrt 2}$$ because my teacher never told use about that.

Also, when do I have to use this: 2 sin A * cos A = Sin 2A ? In what type of problems is this used?
thanks again

The double angle shows up in trajectory type problems when you try to calculate the range of a projectile. We didn't have to invoke it for the problem you posed.

 

[mad]

The double angle shows up in trajectory type problems when you try to calculate the range of a projectile. We didn't have to invoke it for the problem you posed.

I was doing another problem and I had to use this. It was
(9.8 * 53) / 2* 87^2 = sinX* cosX

so I did this:
(9.8 * 53)/ 87^2 = 2*sinX*cosX
-->
(9.8 * 53)/ 87^2 = Sin 2X

but I have no idea how to find this.. the teacher never talked about it. Ill try to find out on the internet. I just want to thank you again for all those replies and help.