a)the stone moves with equal slowness:
V=V0−gt
V0=u
V=u−gt
the maximum lifting height occurs when the stone begins to move down
V=0
u−gt=0
t=gu
height of the stone above the ground:
h=V0t−2gt2=ut−2gt2
hmax when t=gu
hmax=u∗gu−2g∗g2u2=2gu2
b) dimensional analysis:
hmax=2gu2
hmax−[m]
u−[m/s]
g−[m/s2]
2gu2−[m/s2m2/s2]=[m]
the dimensions of the left and right sides
of the formula coincide
Answer:hmax=2gu2
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