A particle is projected vertically upward with a velocity of 15 m/s from the point O. Find
the:
(i) maximum height reached.
(ii) time taken for it to return to O.
(iii) time taken for it to be 25 m below O.
1
Expert's answer
2020-06-28T18:51:44-0400
According to Newton's second law
mdtdv=F
By the Law of conservation of energy
K+P=const
Let y denote the height of the particle above the ground.
Let v(t) denote the velocity of the particle at time t
2mv02+0=2mv2+mgy
mdtdv=−mg
v(t)=v0−gt
(i) maximum height reached.
v=0,y=ymax
2mv02+0=0+mgymax
ymax=2gv02
ymax=2(9.8m/s2)(15m/s)2≈11.480m
(ii) Let t2= time taken for it to return to 0.
When the partice reaches maximum height
v(t1)=0=v0−gt1
t1=gv0
t2=2t1=g2v0
t2=9.8m/s22(15m/s)≈3.061s
(iii) time taken for it to be 25 m below O.
v(t)=dtdy
y(0)=0,v(t)=v0−gt
y=∫v(t)dt=∫(v0−gt)dt=y(0)+v0t−2gt2==v0t−2gt2
Given y=−25m
−25=15t3−29.8t32,t3≥0
4.9t32−15t3−25=0
t3=2(4.9)15±(−15)2−4(4.9)(25)=
=9.815±715
Since t3≥0, we take
t3=9.815+715≈4.259s
The particle will be 25 m below approximately 4.259 seconds after start.
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