A ball is thrown up and attains a maximum height of 19.6m.Its initial speed was
Answers
Answered by
11
Height=h=19.6m
g=-9.8m/s^2(For Upward)
Final velocity=vf=0 m/s
Inital velocity=vi=?
Solution:
Using 3rd Eq of motion
2gh=vf^2-vi^2
2(-9.8)(19.6)=0-vi^2
vi^2=384.16
vi= (384.16)^1/2
vi=19.6m/s
g=-9.8m/s^2(For Upward)
Final velocity=vf=0 m/s
Inital velocity=vi=?
Solution:
Using 3rd Eq of motion
2gh=vf^2-vi^2
2(-9.8)(19.6)=0-vi^2
vi^2=384.16
vi= (384.16)^1/2
vi=19.6m/s
abhishekprajapati14:
thanks
Answered by
5
H=us quare/2g
19.6=us quare/2(9.8)
19.6×19.6=usquare
u=19.6
19.6=us quare/2(9.8)
19.6×19.6=usquare
u=19.6
Similar questions