Question

A particle dropped from the top of a tower h high and at the same moment another particle is projected upwards form the bottom. they meet when the upper one has covered (h/n) of the distance. show that the velocities when they meet are in the ratio 2:(n-2) and its initial velocity is √(ngh/2).​

Answer 1

Answer:

So, you know that you're dealing with two particles, one at the top of a tower that's h meters high, and the other one on the bottom.

Moreover, you know that if you drop the top particle with no initial velocity and launch the bottom particle with an initial velocity equalto v0, they meet when the upper particle travelled 1/n of the heigt of the tower.

For the upper particle, the distance it travelled can be written as

1/n*h = v0upper = 0*t+1/2*g*t^2

1/n*h = 1/2*g*t^2  ---------------------------(1), where

t - the time needed for the particle to reach that height.

The bottom particle will travel a distance of:

h−1/n*h = (n−1)/n*h

For the bottom particle you can write

(n−1)/n*h = v0*t−1/2*g*t^2----------------------------(2)

Since the two particles meet up at this point, the time of flight for both particles must be equal.

Use equation (1) to find a relationship for t^2

t^2 = 2h/ng

Putting this into equation (2) to get

(n−1)/n*h = v0*√2h/ng−1/2 * g * 2h/ng

(n−1)/n*h = v0*√2h/ng − h/n

Rearrange to isolate v0 on one side of the equation

v0*√2h/ng = (n−1)/n*h + h/n

v0*√2h/ng = n/n * h − h/n + h/n

v0 = h / √2h/ng = h * √2h/ng / 2h/ng

    = ng/2 * √2h/ng

This is equivalent to

v0 = ⎷n^2*g^2/4 * 2h/(n*g)

v0 = √ngh/2

Explanation: