Can you solve the Circled Question. I am giving the person 25 points!
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hope u understand the calculations yourself
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KarthikBonthala:
how
tq
when she loses the contact her velocity was 6 nw she moves under gravity alone so using eq. of motion under gravity....at highest pt sh will b at rest and will fall back so highest pt velocity is zero
and acc due to gravity hs 10 as its value
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K.E.=1/2mv^2
m=64 kg
v=6m/s
K.E.=1/2×64×6×6
=1152 J
P.E. at the ground=0
K.E. at the ground=1152 J
so, K.E. at highest point=0
so,P.E. at the highest point=1152 (law of conservation of energy)
mgh=64×10×h=1152 J (here taking g=10m/s^2)
so,h=1.8m
m=64 kg
v=6m/s
K.E.=1/2×64×6×6
=1152 J
P.E. at the ground=0
K.E. at the ground=1152 J
so, K.E. at highest point=0
so,P.E. at the highest point=1152 (law of conservation of energy)
mgh=64×10×h=1152 J (here taking g=10m/s^2)
so,h=1.8m
This is the answer I wanted to get.
This is the format, without k.e it not possible for writing
okkkkk you should have said solve the problem using concept of law of conservation of energy
Sry but you should understand it right
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