Question

small Steel balls Falls from rest through the opening at A at the steady rate of n balls per second find the vertical separation h of two consecutive balls when the lower one has dropped d meters

Answer 1

Answer:

S1 = 1/2 g t1^2     distance first ball falls

S2 - 1/2 g t2^2 = 1/2 g (t1 -1)^2 = 1/2 g (t1^2 - 2 t1 + 1)

h = S1 - S2 = 1/2 g (2 t1 - 1)

d = S1 = 1/2 g t1^2

t1 =  (2 d / g)^1/2     time for first one to fall d

h = 1/2 g [2 *  (2 d / g)^1/2 - 1]

Answer 2

The correct question: Small steel balls fall from rest through the opening at a steady rate of 2 balls per second. The vertical separation "h" of two balls when the lower one has dropped 20 meter is?

The vertical separation "h" of two consecutive balls is calculated as follows.

Given:

The height from which the balls dropped, h₁ = 20 m.

Acceleration due to gravity, g = 10 m/s.

To Find:

Using given data, we have to find the vertical separation "h" of two consecutive balls when the lower one has dropped 20 meter.

Solution:

Equation for the height from which the balls dropped, "h₁", is given by,

$h_1$  $= \frac{1}{2} .g.t^{2}$

Substituting values for "h₁" and "g", we can find out the the time taken to cover 20m.

$20 = \frac{1}{2} .10.t^{2}$

$t^{2} =$ $20.(\frac{2}{10} )$

$i.e., t = 2.$

The second ball is thrown after 1 second.

∴, The distance (s) covered by the second ball in (2-1) seconds is,

$s = \frac{1}{2} × 5 × (1) = 2.5 m$

∴, The vertical separation = $20 - 2.5 = 17.5 m.$

Hence the vertical separation "h" of two consecutive balls when the lower one has dropped 20 meter is 17.5 m.