Asked by utsman
A ball thrown vertically upwards from ground level hits the ground after 4 seconds.Calculate the maximum height it reached during its journey.
Answers
Answered by
Henry
Tf = 4/2 = 2 s. = Fall time.
h = 0.5g*Tf^2 = 4.9 * 2^2 = 19.6 m.
h = 0.5g*Tf^2 = 4.9 * 2^2 = 19.6 m.
Answered by
Favor
Good
Answered by
SAMUEL
DANIEL
Answered by
AISHAH
intial velocity T = 2U/G 4 = 2U/10 = 4=U/5 U = 4 X 5 U = 20 M maximum height H = U^2/2G H = 20^2/2 X 10 H=2OM
Answered by
Abah Derek Abah
s= ut +1/2at^2 , but t=4/2 = 2s , u=0 , a= 10m/s therefore, s= 0×2 + (10× 2^2)/2
s=(10×4)/2
s=40/2 = 20
s=(10×4)/2
s=40/2 = 20
Answered by
Oche
Superb
Answered by
Oche
T=2U/g 4=2U/10=U20 From h=u2/2g =400/20 H=20m
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