Initial Velocity of a projectile, how?

[free-node-5]

Initial Velocity of a projectile, how?

If I have a problem that says the x and y distance of a target from a starting position, and only gives the angle at which the particle is launched, how do I find the required initial Velocity for its launch?



I've been given the following equations to work with:
V_f = V_i + at
x_f = x_i + V_it + (.5)at²
V_f² = V_i² + 2ad



I've spent hours trying to find a solution like this and can't seem to get anywhere. At one point I wrote out a huge equation with a bunch of trig stuff and it said a human threw a ball at over 513m/s so I know it must have been wrong.
One line I have written in my work says:
1.05=( V_isin43 )/( V_icos43 ) - ( 4.9 * 9 )/( V_i cos 43 )
but I'm just asking in general, how do you do this?

 

[kuruman]

If you are given the x and y distance, in general, you use the first equation to solve for t in terms of x then plug that in the second equation, That gives you y as a function of x with the initial speed and angle as parameters. Use the third equation to find the maximum height reached by the projectile, in general.

 

[free-node-5]

I took what you said to heart and I still seem to just be making a mess.

Here are a few lines from my work:
1.05 = V_y(x_f/Y_ix) - 4.9 ( x_f² / V_ix² )
...
1.05 = 9 tan43 - 4.9 ( 80 / V_ix² )
...
V_ix = .136 ?
...
t = x_f / V_ix
t = 9 / .136 = 66.176

seems like a long time for something to get to a target that close :-/


Do you think you could show me or maybe just give an example?
My values are $$\theta$$=43(degrees), x_i = 0, y_i = 0, x_f = 9, y_f = 1.05

thanks

 

[kuruman]

Can you state the problem exactly the way it is given? Your interpretation of the given quantities may be in error.

 

[free-node-5]

quoting is probably not a good idea but trust me they were pretty straight-forward
There was even a diagram below the problem

--
A guy who's 2 meters tall, stands 9 meters horizontally from a ball-hoop mounted 3.05m high.
He throws the ball at an angle of 43 degrees from the horizontal. How hard must he through the ball (m/s) for it to go into the hoop but not hit the backboard?
--

 

[rl.bhat]

I took what you said to heart and I still seem to just be making a mess.

Here are a few lines from my work:
1.05 = V_y(x_f/Y_ix) - 4.9 ( x_f² / V_ix² )
...
1.05 = 9 tan43 - 4.9 ( 80 / V_ix² )
...
V_ix = .136 ?

This equation is correct.
But V_ix is wrong. Check the calculation.

 

[free-node-5]

extreme thanks

I'm not sure what I did at this point because I failed so many times but when i fixed that mistake, I managed to get to a correct final solution