An Olympic basketball player shoots towards a basket that is 5.98 m horizontally from her and 3.05 m above the floor. The ball leaves her hand 1.90 m above the floor at an angle of 60.0o above the horizontal. What initial speed should she give the ball so that it reaches the basket and hopefully scores?

Question

Been using t= 5.98/(cos60) Vo and plugging it into
3.05 = 1.9+ (Sin 60 vo)t+ 1/ (-9.81) t^2 and getting a different answer everytime
The wrong answerss are
1 Incorrect. (Try 1) 5.17 m/s
2 Incorrect. (Try 2) 11.6 m/s
3 Incorrect. (Try 3) 1.14 m/s
4 Incorrect. (Try 4) 5.64 m/s
>...< Please. Someone help

Answer 1

vertical:
3.05=1.90+Visin60*t-4.9t^2
horizontal:
5.98=Vicos60*t or t= 5.98*2/Vi

putting that into vertical
3.05=1.90+ViSin60*2*5.98/Vi-4.9(2*5.98/Vi)^2

which is not your equation. Solve this for Vi, notice it is a quadratic, get it into standard form, and use the quadratic equation.