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...

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# Basketball question HELPP

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