1. A stone is dropped from top of a tower. When it crosses a point 5 m below the top , another stone is dropped from a point 25 m below the top. Both stones reach the bottom of the tower simultaneously. Find the height of the tower. Take g = 10 m/s2
2. A particle is projected from ground with a velocity of 40 m/s at 60º with the horizontal. Find the times at which the velocity of the particle makes an angle of 45º with the horizontal.
3. A particle is projected at an angle 60º with horizontal with a speed of 20 m/s . Taking g = 10 m/s2 , find the time after which the speed of the particle remains half of it's initial speed.
Yes, I am weak in Mechanics , and these may appear to be simple problems to many.
Knowing that, I request helpful people to help me out. Will be really obliged if you write the proper statements for each step ( very concisely ) , instead of just the mathematical calculation.
Thanks . :)
71Q1:
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.
71Q2:
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: Uy = U sin 60° and Ux = U cos 60°
Finally: Vy = V sin 45°= U sin 60° - gt22 and Vx = V cos 45°
From these, We must have U[x] = V[x].
You can solve.
11after t sec. the the x component of the velocity will remain same..... that is Vx=20sin60°.
now the Vy will only change..........
now... after time say t sec. the
V'y=Vy-gt.....
now after t sec. the the resultant velocity is said to be half of the initial velocity...
or
Vt=√V'2y+Vx2=2Vi...
Now...solve the equation and get the value of t..............