1.

two bodies A and B are thrown upwards with 40 metre per second and 80 metres per second respectively. the ratio of the distances covered by them in last second(of their vertical journey)

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Answer

  • 1 : 1

Given

  • Two bodies A and B are thrown upwards with 40 metre per second and 80 metres per second respectively

To Find

  • The RATIO of the distances covered by them in LAST second(of their vertical journey)

SOLUTION

\rm Let\ ,u_1=40\ m/s \;\; \& \;\; u_2=80\ m/s

The final VELOCITIES of both bodies A and B be " 0 " Since , at top position they both be at rest .

So , v₁ = 0 m/s & v₂ = 0 m/s

g MUST be -ve as we throw both balls against the gravity .

Now , apply 1st equation of motion .

\implies \rm v_1=u_1+(-g)t_1\\\\\implies \rm 0=40-10t_1\\\\\implies \rm 10t_1=40\\\\\implies \rm t_1=4\ s

Likewise ,

\rm v_2=u_2+(-g)t_2\\\\\implies \rm 0=80-10t_2\\\\\implies \rm 10t_2=80\\\\\implies \rm t_2=8\ s

We know that ,

\rm S_n=u+\dfrac{a}{2}(2n-1)

Now we need to find the ratio of the distances at last second .

\implies \rm \dfrac{s_4}{s_8}\\\\\implies \rm \dfrac{u_1+\dfrac{-g}{2}(2(4)-1)}{u_2+\dfrac{-g}{2}(2(8)-1)}\\\\\implies \rm \dfrac{40+\dfrac{-10}{2}(8-1)}{80+\dfrac{-10}{2}(16-1)}\\\\\implies \rm \dfrac{40-5 (7)}{80-5(15)}\\\\\implies \rm \dfrac{40-35}{80-75}\\\\\implies \rm \dfrac{5}{5}\\\\\implies \rm \dfrac{1}{1}

So , The ratio of the distances covered by them in last second is S₁ : S₂ = 1 : 1


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