sha30
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
rohit4510
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
Answers & Comments
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