Question

a stone is thrown vertically upward with velocity 25.5 m/s .the distance travelled by the particle in last second of its upward motion is ( take g=10m/s)

Answer 1

hi,

u=25.5m/s

v=0

a=g=10m/s2

by using the equation.v=u+at

0=25.5-10t

t=2.55

now,Sn=un+1/2a(2n-1)

here n=2.55,a=10,u=25.5

Sn=25.5*2.55+10/2(approx 11)

   =120.o25m is ANSWER

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