Review Q #5 (Projectile Motion)

[TeenieWeenie]

Homework Statement

A basketball leaves the hands of a shooter 2.1 meters above the floor and soars towards the basket, which is at a horizontal distance of 3 meter from the shooter. If the basket is 2.6 meters above the floor, and the player shoots the ball at an angle of 45 degree above the horizontal, calculate the initial velocity the shooter must give the basketball in order to score a basket.

Yo = 2.1m
Xo = 3m

Y = 2.6m
Ao = 45 degrees above horizontal (?)

Homework Equations

Vx = VoCosAo
Vy = VoSinAo - gt

The Attempt at a Solution

I don't understand what the X is supposed to be?

 

[TeenieWeenie]

Ok, so I've drawn the diagram for this.
It looks like this is solved with trig?
I get a right triangle with the top as 2.6m. and the lower 45 degree angle as 2.1m.
So the height of the triangle would be 0.5m and the base would be 3m.

Then, what does the hypotenuse equal?

 

[Thaakisfox]

This is projectile motion. No it cannot be solved with trig. You have to use physics too ;)

 

[TeenieWeenie]

Hehe.
What am I missing :( ?
I have no idea where to go from there (I've been looking at this for 10 minutes straight), yet again.

 

[Thaakisfox]

First you know that the horizontal distance x=Vo*cos(45)*t.
And you know the vertical distance: y=Vo*sin(45)*t-(g/2)t^2.
Now from the first equation take t and plug it into the second equation. You will get an equation for y in terms of x.
You want that y=0.5 and x=3 right? as the total height difference between the point of throw and the height of the basket is (2.6-2.1=0.5).

 

[TeenieWeenie]

3 = Vo * cos45^o * t
t = 3/ (Vo * cos45^o)

0.5 = Vosin45^o-(9.8/2)t²
0.5 = 3 tan45 - (9.8/2)(9/cos²45^oV_o²

V_o²= [(3tan45^o-0.5)/(9.8/2)]*(9/cos²45^o)

V_o²= (2.5/(9.8/2) * (0.5/9)
V_o²= 0.2834467
V = (0.2834467)^1/2 ?

That's...odd?
I don't think that's right...

 

[Thaakisfox]

You messed up the calculation. Check that you expressed the velocity correctly.

 

[TeenieWeenie]

Alrighty, looks like I had computational error.

0.5 = 3 -(9.8*9/(Vo^2*cos^2(45))/2
11.025/ [Vo^2cos^2(45)]= 2.5
11.025/2.5 = Vo^2 * cos^2(45)
4.41 = [Vo^2]/2
8.82 = Vo^2
2.9698 m/s = Vo

how about that?

 

[Thaakisfox]

That still looks wrong ;)

 

[TeenieWeenie]

What could be it!? :(

 

[Thaakisfox]

Check from here again :
0.5 = 3 -(9.8*9/(Vo^2*cos^2(45))/2
and calculate step by step