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?(adsbygoogle = window.adsbygoogle || []).push({});

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...

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Basketball question HELPP

**Physics Forums | Science Articles, Homework Help, Discussion**