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if a particle takes t seconds less and acquires a velocity of vm/s more in falling through the same distance on two planets whe accelarations due to gravity are 2g and 8g respectively the what is the value of v?

if a particle takes t seconds less and acquires a velocity of vm/s more in falling through the same distance on two planets whe accelarations due to gravity are 2g and 8g respectively the what is the value of v?

Grade:12

3 Answers

Chetan Mandayam Nayakar
312 Points
9 years ago

let u1(t) and u2(t) be the speeds on 2g and 8g planets respectively

let motion start at t0,fall start at T,and T1(8g), and falls end at t1, and t2(8g)

let fall length be d,

u1(T)=2gT, u2(T)=8gT,u1(t1)=2g(t1),and u2(t2)=8g(t2)

d=(u1(T)+u1(t1))/(2(t1-T)

d=(u2(T1)+u2(t2))/(2(t2-T1)

t2-T1=t

u1(T1)-u2(t2)=v

 

 

 

 

 

goutham 1995 gumm
18 Points
9 years ago

the answer is 4gt. can anyone explain how? 

thank you in advance.

Swapnil Saxena
102 Points
9 years ago

U v to take only 3 equations in consideration. Here r d following

Assume v0 be the velocity achieved by the particle in 2g . In 8g it will achive a velocity of v0+v 

Assume that the particle take t0 time in 8g. Then it had took a time of t+t0 in 8g.

Using Newtons first eq.

 v0 = 2g(t+  t0) ---------(I)

(v+ v0)= 8gt -------(II)

Solving the pair of equation gives v=6g(t-t0)  ---(III)

Using Newtons second equation .

Assume the height through which they fall = H

H =1/2(2g)( t+t0)^2 and H= 1/2(8g) (t0)^2 ==> 1/2(2g)( t+t0)^2 = 1/2(8g) (t0)^2

 ==> (t0)= t/3 

Substituting value of  (t0) in eq III gives

v= 6g(2t/3) = 4gt

 

 

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