A stone is thrown vertically upward with an initial velocity v . The distance travelled by it in time 1.5v/g?

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Answer 1 · 2017-07-24T04:17:02

The initial upward velocity of the stone #=v#

If time required to reach at its maximum height #H# be #t_1# then
#0=v-g*t_1#

So #t_1=v/g#

Now #0^2=v^2-2gH#

So #H=v^2/(2g)#

So during time #t_1# the stone will traverse a distance #H=v^2/(2g)=(0.5v^2)/g#

The remaining time #t_2=(1.5v)/g-v/g=(0.5v)/g#

During next phase of its downward journey it srarts with velocity #=0# and after time #(t_2=(0.5v)/g)# its displacement will be

#h=0*t_2+1/2g t_2^2#

#=>h=1/2g ((0.5v)/g)^2#

#=>h=((0.125v^2)/g)#

Hence total distance covered during the given time #(1.5v)/g# will be

#=H+h#

#=(0.5v^2)/g+(0.125v^2)/g#

#=(0.625v^2)/g#