A ball is thrown vertically upward attains a maximum height of 45 m . The time after which velocity of the ball becomes equal to half the velocity of projection?
Nidhi2002:
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v^2 =u^2+2as
0^2=u^2+2*(-10)*45
u^2=900
u= √900
u=30
taking v as 15,
we get
v=u+at
15=30+(-10*t)
15=30-10t
t=15/10= 3/2=1.5m/s
0^2=u^2+2*(-10)*45
u^2=900
u= √900
u=30
taking v as 15,
we get
v=u+at
15=30+(-10*t)
15=30-10t
t=15/10= 3/2=1.5m/s
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