Question

two bodies are projected vertically from the mid point of a tower with equal speeds so that one just reached the top of the tower the ratio of times of flight of the teo bodies is

Answer 1

isn't it 1? or i don't get the question?

Answer 2

the ans=3 + 2root2

Answer 3

But how? two bodies go with same velocity from the same point.how could the time periods be different?

Answer 4

one is going up and other is comming down from the mid point of the tower

Answer 5

i think you did not mention that in the question :)

Answer 6

Sorry shameer1. My internet went out earlier when I was helping you.

Let's set up the equations of motion for the two bodies. Positive will be upwards. The body thrown upwards will be denoted by subscript u and the body thrown downwards with subscript d. \[y_u(t) = d_i + vt_u - {1 \over 2} g t_u^2\]\[y_d(t) = d_i - v t_d - {1 \over 2} g t_d^2\]We need to find the time such that y=0. Solving time using the quadratic equation yields. \[t_u = {v \pm \sqrt{v^2 + 2d_i g} \over g}\]\[t_d = {-v \pm \sqrt{v^2 + 2d_ig} \over g}\]We can ignore the negative solution.\[t_u = {v + \sqrt{v^2+2d_ig} \over g}\]\[t_d = {-v + \sqrt{v^2 + 2d_ig} \over g}\]Taking the ratio\[{t_u \over t_d} = {v + \sqrt{v^2+2d_ig} \over -v + \sqrt{v^2 + 2d_ig}}\]