Top of a tower a stone, Physics

From top of a tower a stone is thrown up and it reaches the ground in time t1. A second stone is thrown down with the same speed and it reaches the ground in t2. A third stone is released from rest and it reaches the ground in time t3 then t3= ?

solution)

let the intial speed be ''u'' and height be ''h''.

h(-j)=u(j)*t1+0.5g(-j)*t12

Case(2)

h(-j)=u(-j)*t2+0.5g(-j)*t22

Case(3)

h(-j)=0.5g(-j)*t32

 

solve case 1 and case 2 for t1 and t2 (quadratic) .

 

Posted Date: 3/26/2013 12:57:50 AM | Location : United States





its very easy quest. we just need to find the values in u,g,and v(final velocity,this is constant for the first two cases)

for first throw.

t1= 2u/g + v-u/g=v+u/g

for second throw.

t2=v-u/g

for third release.

t3=u/g

so t3=(t1-t2)/2g

 

Posted by | Posted Date: 3/26/2013 12:58:51 AM


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