Question

a particle is dropped from a point A at a certain height from ground. it falls freely and passes through points B,C and D with BC=CD. time taken to move from A toB is​

Answer 1

v = u + a*t

Initial Velocity u m/s

Final Velocity v m/s

Time t/s

Acceleration a m/s²

ANSWER

BC=2V+

2

g

×2

2

BC=2V+2g

2BC=3V+

2

1

g×3

2

2BC=3V+

2

9g

4V+4g=3V+

2

9g

V=

2

9g

−4g

=g/2

A→B

V=u+at

2

g

=0+g×t

t=1/2

Answer 2

Answer:

$t=0.5 s$

Explanation:

Let velocity of the particle at point B be v.

Now,$\quad B C=2 v+\frac{1}{2} g \times(2)^{2}$

or,$\quad B C=2 v+2 g$..........(i)

Similarly, $\quad 2 B C=3 v+\frac{1}{2} \times g \times(3)^{2}$

$\Rightarrow \quad 2 B C=3 v+\frac{9 g}{2}$.......(ii)

From Eq. (i),

$\begin{aligned}2(2 v+2 g) &=3 v+\frac{9 g}{2} \\\Rightarrow \quad 4 v+4 g &=3 v+\frac{9 g}{2}\end{aligned}$

$\begin{array}{ll}\Rightarrow & 4 v-3 v=\frac{9 g}{2}-4 g \\\Rightarrow & v=\frac{g}{2}\end{array}$...........(ii)

From point A to B,

$& \Rightarrow & v &=u+a t \\\Rightarrow & & \frac{g}{2} &=0+g \times t \\\Rightarrow & t &=\frac{1}{2} \mathrm{~s} \\& \Rightarrow & t &=0.5 \mathrm{~s}$