Question

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)

Answer 1

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

Answer 2

$\bigstar$ $\sf{\purple{\underline{\underline{\blue{To\:Find:-}}}}}$

• The ratio of the distances covered by them in last second .

$\bigstar$ $\sf{\blue{\underline{\underline{\purple{SOLUTION:-}}}}}$

☞ GIVEN :-

• Body “A” is thrown upwards with velocity “40 m/s”

• Body “B” is thrown upwards with velocity “80 m/s”

☞ CONCEPT :-

• When body moving with constant acceleration, then Equation of Motion is,

$\checkmark\:\tt{\underline{\green{\boxed{\:v\:=\:u\:+\:at\:}}}}$

☞ Where,

• v = final velocity
• u = initial Velocity
• a = acceleration
• t = time

☞ Here, acceleration acts on downward direction and gravitational acceleration is applied on both the body .

• a (acceleration) = - 10 m/s²

☞ CALCULATION :-

☞ First calculate, the time required in whole journey for both bodies .

♡ Case - 1 :- For body “A” ,

Here,

• v = 0 m/s
• u = 40 m/s
• a = - 10 m/s²

$\tt{\implies\:0\:=\:40\:+\:(-10)t\:}$

$\tt{\implies\:(-10)t\:=\:-40\:}$

$\tt{\boxed{\implies\:t_A\:=\:4\:s\:}}$

♡ Case - 2 :- For body “B” ,

Here,

• v = 0 m/s
• u = 80 m/s
• a = -10 m/s²

$\tt{\implies\:0\:=\:80\:+\:(-10)t\:}$

$\tt{\boxed{\implies\:t_B\:=\:8\:s\:}}$

☞ FORMULA :-

$\checkmark\:\tt{\underline{\purple{\boxed{s^{n\:th}\:=\:u\:+\:{\dfrac{1}{2}}\:a\:(2n\:-\:1)\:}}}}$

• $\tt{s^{n\:th}\:→\:distance\:or\:displacement\:in\:(n^{th})\:second\:}$

☞ Now, put the above formula and get the Answer .

$\tt{\implies\:{\dfrac{s^{4th}}{s^{8th}}}\:=\:{\dfrac{40\:+\:{\dfrac{1}{2}}(-10)\:(8-1)}{80\:+\:{\dfrac{1}{2}}(-10)\:(16-1)\:}}\:}$

$\tt{\implies\:{\dfrac{s^{4th}}{s^{8th}}}\:=\:{\dfrac{40\:-\:35}{80\:-\:75}}\:}$

$\tt{\implies\:{\dfrac{s^{4th}}{s^{8th}}}\:=\:{\dfrac{5}{5}}\:}$

$\tt{\boxed{\implies\:{\dfrac{s^{4th}}{s^{8th}}}\:=\:{\dfrac{1}{1}}\:}}$

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::\:The\:ratio\:of\:the\:distance\:covered\:by\:both\:body\:is\:\:(1\::\:1)\:}}}}$