Question

A particle is projected from the top of a tower upward and it takes time t1 to hit the ground

Answer 1

Explanation:

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.

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) .

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

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)

In case there u=0

So S= 0.5*g*t3*t3…………………………………………(2)

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

Solution : t3=sqrt(t1*t2)

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