consider that the final veloicty is the negative of the intial velocity...
Vf=Vi+gt
2Vi=gt
vi=1/2 gt solve for vi
h= vit-1/2 g t^2 and use t=3.5/2 for time to get to max height.
a guy throws a ball up and catches it 3.5 seconds later. what is the max height and initial velocity.
Please help!
4 answers
4 meters
4 meters 8 m/s
17.16m/s and the displacement is 15meters
1 answer