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# 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= ?

Akash Kumar Dutta
98 Points
8 years ago

Dear Abhijeeth,

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. (ANS)

Regards.

vijay reddiar
30 Points
8 years ago

Case(1)

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) .

raghul ravindran
14 Points
8 years ago
1. In the first case stone is thrown upwards. So first it goes up against gravity and then comes down. So its motion from going up to coming down to original throwing position is a projectile motion.

Applying equations of motion:

V= u+at

Let u= u1

a = g = 9.8m/sec2

We divide t1 into two timings , one for projectile motion and second part a stone travels from original position to ground.

So, t1=t’+t”

t’ = 2u1/g

t” = t2 as after stones reaches back original position it has velocity u1.

t1=2u1/g+t2

1. In case 2 stone is thrown down with u1 velocity. Time taken is t2. Let us say distance covered by stone before reaching ground be S.

S= u1*t2+0.5*g*t2*t2  …………………………………(1)

1. In case there u=0

So S= 0.5*g*t3*t3…………………………………………(2)

1. Equating (1) & (2) and multiplying 2/(g*t2) on both sides we have:

Solution : t3=sqrt(t1*t2)

Dablu kumar
13 Points
2 years ago
1. A body is thrown up from a tower. If the ratio of distance to displacement is n, the ratio of time for upward and downward motion is
(a)                           (b)                         (c)                    (d)