Some physics problems i need solution

Q1:

Let the time taken by the stone released from a point 25m below top to reach ground is ' t ' seconds. It is the same time taken by the other stone to reach the ground from a point 5 m below the top of tower.

So, Assuming ' h ' to be the height of the tower we write two kinematical equations for the two blocks as :

h-25= gt²2 .... (i)

and

h-5 = 10t + gt²2 .... (ii) [ 10m/s in the velocity of the stone dropped from top of tower when it has descended 5m]

Solve these for t and and then h.

Q2:

We know that velocity of particle won't change in horizontal direction. So we'll use that Idea.

Let after time t, the velocity of the particle will make 45° with horizontal.

Initially: U_y = U sin 60° and U_x = U cos 60°

Finally: V_y = V sin 45°= U sin 60° - gt²2 and V_x = V cos 45°

From these, We must have U[x] = V[x].

You can solve.

after t sec. the the x component of the velocity will remain same..... that is V_x=20sin60°.
now the V_y will only change..........
now... after time say t sec. the
V'_y=V_y-gt.....

now after t sec. the the resultant velocity is said to be half of the initial velocity...
or
V_t=√V'²_y+V_x²=2V_i...

Now...solve the equation and get the value of t..............