1. Oct 11, 2006

### dragos

A basketball player shoots a ball at an angle of 60 with an initial velocity of 7.0 m/s. The ball was 2.0m above the ground when released and the basket was 3.00m high. How long was the ball in the air?

Here is what i did so far:
y-axis x-axis
v1=7.0m/s x sin60 =6.06 m/s v1=7.0m/s x sin30 = 3.5m/s
a=-9.81m/s2

I used v2squared = v1squared +2ad
since v2=0 0=6.06m/s squared + 2(-9.81m/s2)d
d=1.87

To find time i used d=vt+1/2atsquared
-1.87m=(6.06m/s)2t + 1/2(-9.81m/s2)t2
after doing the quadratic formula i got t=1.49113
Therefore the ball was in the air for 1.49113s/2 = 0.75s
Should i divide it by 2 or noooo??? im not sure if its the maximum height...

2. Oct 11, 2006

### quasar987

Why would the y-velocity be 0 when the ball reaches the net?

3. Oct 11, 2006

### dragos

v2

so v2 isn't zero in this case. i was supposing that v2 will be 0, when the ball hits the ground. So how would you proceed with this problem further?

4. Oct 11, 2006

### quasar987

I would write the equation of kinematics for y(t) and solve for the time when y=3. I will have to solve a quadratic equation in t. The lesser solution for t is the first time the ball got up to 3m (it hapened then the ball as ascending). The greater solution is what I'm looking for; it is the time when the ball is descending and on its way to a glorious 3-pointer.

5. Oct 11, 2006

### drpizza

I'd think that your second equation is sufficient, except you'll want to "fix" it to either:

$$d=d_0 + v_0t+\frac{1}{2}at^2$$

or else, if you're not familiar with having a starting position on the right side of that equation, simply use the vertical displacement the ball is going to have: 1.00 meters.

Of course, you'll use the vertical component of velocity that you already found. Incidentally, I believe there will be two solutions for t; one corresponding to shooting up through the hoop, and the other for it falling down through the hoop.