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Grade 12th passMechanics

A person is throwing balls into the air one after the other. He throws the second ball when first ball is at the highest point. If he throwing 2 balls every second, how high do they rise?

Profile image of Bincy
10 Years agoGrade 12th pass
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3 Answers

Profile image of ram charan
10 Years ago
v=0
time taken by each to reach hmax=1/2s
0=u-g/2
u=g/2
0=u^2-2gh
u^2=2gh
g^2/4=2gh
h=g^2/8g
h=2g/8
Profile image of Ajit
10 Years ago
Time taken to reach maximum height - t= u\g
Therefore 1/2= u/ 20
Ten u= 5
Max height = u^2/2g
That is 5^2/20 = 1.25 m
Profile image of Vandana Agrawal
9 Years ago
1 ball is taking 2s to travel from initial position to the final position.T=2u/g2=2u/10u=10m/sS=u*u/2gS=100/20S=5m