Recent content by jkcali2ooo

Thanks for the help Berkeman. When I calculate out time and re-plug that back in with the inital velocity I calculated, I would get to the 1.35m (a.k.a. the hoop.) However, I was wondering if I can just reduce the height where the ball was initially shot from and then just subtract that out...

Sorry I added another Vi. So 1.35 = (0.866025Vi)(6.1/Vi) - 4.9(6.1/Vi)^2 which then leads to... 1.35 = 5.282754 - 182.329 / Vi^2 So the Vi's cancel each other out so I don't have to use the quadratic forumula. I wasn't sure about this step either cause it seemed weird that they...

I believe that intial velocity is unknown since the ball is already traveling at some speed in m/s. However, at the max height of the ball when it reaches the hoop would have a velocity of 0 with a negative accelleration of gravity or -9.8m/s^2. I have revised my attempt at the problem...

Homework Statement A basketball player shoots the ball at a height of 1.7m at an angle of 60 degrees into a hoop. The range is 10ft or 3.05m. The height from the ground to the top of the hoop is 10ft or 3.05m. What is the initial velocity m/s to make it to the hoop? Homework Equations...