A basketball player who is 2.00[m] tall is standing on the floor 10.0 [m] from the basket, as in the figure. If he shoots the ball at 40.0degrees angle with the horizontal, at what initial speed must he throw so that it goes through the hoob without striking the backboard? The basket height is 3.05[m].
xf = xi + vit + 1/2at^2
yf= yi +vit + 1/2at^2
The Attempt at a Solution
I used the yf equation to come up with an equation that has vi and t as the variables. Then I used the xf equation and solved for t. Then I plugged in that t into the yf equation. It comes out to a pretty large quadratic equation:
-4.9 (10/vi(sin40))^2 + (vi(cos40))(10/vi(sin40)) - 1.05 = 0
Now I'm having trouble finding vi. I'm pretty sure I did everything right, but can someone help me figure out how to solve for vi?
Also, how do you make it so the quations come up looking normal instead of typing it here, it's formatted different? Is that a website you type it in and link here? Thanks!