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A ball is dropped from rest vertically from a distance of 1.0m and rebounds to a distance of 0.67m.calculate its velocity before it rebounds and its velocity jxt after it rebounds. A ball is dropped from rest vertically from a distance of 1.0m and rebounds to a distance of 0.67m.calculate its velocity before it rebounds and its velocity jxt after it rebounds.

Before it rebounds. . .
Using the formula:
V^2=U^2+2gS

U=O
g=10m/s^2
S=1.0m
v=?
V^2 = u^2+2gS
=2×10×1
=20
Therefore,
V=√20
I just did it online... No rough work, so If I'm wrong, kindly correct me.

ERUDITEE:
Before it rebounds. . .
Using the formula:
V^2=U^2+2gS

U=O
g=10m/s^2
S=1.0m
v=?
V^2 = u^2+2gS
=2×10×1
=20
Therefore,
V=√20
I just did it online... No rough work, so If I'm wrong, kindly correct me.

its correct bt wat of d other velocity

profftrojahn:
its correct bt wat of d other velocity

here's other part-
After Rebound...
V^2=U^2 + 2gh
since the ball rebounds dat means it goes up,thus making it to have accelaration due to gravity (g) to be negative.
g= -10
h=0.67
V=0 (because the ball is going up)
substitute values in d formular...
0=U^2 - 2*10*0.67
U^2=13.4
so U= square root of 13.4
So U=3.66.
Sorry fr any mistake I made in ma language n if am wrong pls tell...cos I knw dis is correct

Grammyz:
here's other part-
After Rebound...
V^2=U^2 + 2gh
since the ball rebounds dat means it goes up,thus making it to have accelaration due to gravity (g) to be negative.
g= -10
h=0.67
V=0 (because the ball is going up)
substitute values in d formular...
0=U^2 - 2*10*0.67
U^2=13.4
so U= square root of 13.4
So U=3.66.
Sorry fr any mistake I made in ma language n if am wrong pls tell...cos I knw dis is correct

its correct