Homework Help: Need help for a projectile problem

1. Sep 27, 2004

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

Last edited: Sep 27, 2004
2. Sep 27, 2004

Tide

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

3. Sep 27, 2004

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?

Thank you very much for your time

4. Sep 27, 2004

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.

5. Sep 27, 2004

Tide

P.s.

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

Last edited: Sep 27, 2004
6. Sep 27, 2004

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

7. Sep 27, 2004

faust9

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

8. Sep 27, 2004

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 couldnt 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

9. Sep 27, 2004

Tide

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.

10. Sep 27, 2004