Question

0.22. (a) A stone is allowed to fall from a tower of height 200 m and at the same time anothenstone is
projected vertically upwards from the ground at a velocity of 20 m/s. Calculate when and where the
stones will meet..
(b) The walls of your classroom are in motion but appear stationary. Explain​

Answer 1

The 2 stones will meet in 10th second.

Explanation:

• Height "h" =  200 m
• The point at which the 2 stones will meet  = a
• Lets one part be x and other part be 200 -x
• The upper ball will have initial velocity 0.

From 2nd equation of motion.

x =ut + at^2

As initial velocity is zero u = 0

x = 1/2 at^2

x = 1/2 gt^2 (a=g)

x = 1/2 ×10× t^2

x = 5 t^2 -------(1)

For lower stone:

u = 20

x = 200 - x

Using second equation of motion.

x = ut +1/2at^2

200 - x = 20× t -1/2gt^2 (a=-g)

200 - x = 20 t -5 t^2 -----------(2)

Substitute equation (1 ) in equation (2):

200 - 5 t^2 = 20 t - 5 t^2

t =10 sec

Substitute t in equation (1)

x =  5× 10×10

x = 500 m

Therefore, the 2 stones will meet in 10th second.

Also learn more

A stone is dropped from a height 300 m and at the same time another stone is projected vertically upwards with a velocity 100 m and find the height when they meet  ?

https://brainly.in/question/4876901

Answer 2

Answer:

• The stones will meet after time (t) 10 seconds.

Given:

Ball " A "

1. u = 0 m/s.
2. a = + g = 10 m/s².
3. Let the ball A cover (200 - x) distance when it meets the ball B.

Ball " B "

1. u = 20 m/s.
2. a = - g = - 10 m/s².
3. Let the ball B cover "x" distance when it meets the ball A.

Explanation:

$\rule{300}{1.5}$

Applying Second Kinematic equation for Ball " A "

$\large\bigstar \; {\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

$\bold{Here}\begin{cases}\text{S Denotes Distance} \\ \text{u denotes initial velocity} \\ \text{a denotes Acceleration} \\ \text{t Denotes time taken}\end{cases}$

Now,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

$\longmapsto \large{\tt 200 - x = 0 \times t + \dfrac{1}{2} gt^2}$

$\longmapsto \large{\tt 200 - x = 0 + \dfrac{1}{2} \times 10 \times t^2}$

$\longmapsto \large{\tt 200 - x = \dfrac{10}{2} \times t^2}$

$\longmapsto \large{\tt 200 - x = \cancel{\dfrac{10}{2}} \times t^2}$

$\longmapsto \large{\tt 200 - x = 5 \times t^2}$

Rearranging,

$\longmapsto \large{\tt 200 = 5 \; t^2 + x}$

$\longmapsto \large{\tt 200 - 5 \; t^2 = x}$

$\longmapsto\large{\tt x = 200 - 5 \; t^2 \; \quad\dfrac{\quad}{}[1]}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Applying Second Kinematic equation for Ball " B "

$\large\bigstar \; {\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

$\bold{Here}\begin{cases}\text{S Denotes Distance} \\ \text{u denotes initial velocity} \\ \text{a denotes Acceleration} \\ \text{t Denotes time taken}\end{cases}$

Now,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

$\longmapsto \large{\tt x = 20 \times t + \dfrac{1}{2} \times - 10 \times t^2}$

$\longmapsto \large{\tt x = 20 \; t - \dfrac{10}{2} \times t^2}$

$\longmapsto\large{\tt x = 20 \; t - \cancel{\dfrac{10}{2}} \times t^2}$

$\longmapsto \large{\tt x = 20 \; t - 5 \; t^2}$

Substituting the Equation [1] value,

$\longmapsto \large{\tt 200 - 5 \; t^2 = 20 \; t - 5 \; t^2}$

Now,

$\longmapsto \large{\tt 200 \cancel{- 5 \; t^2} = 20 \; t \cancel{- 5 \; t^2}}$

$\longmapsto \large{\tt 200 = 20 \; t }$

$\longmapsto\large{\tt t = \dfrac{200}{20}}$

$\longmapsto \large{\tt t = \cancel{\dfrac{200}{20}}}$

$\longmapsto \large{\underline{\boxed{\red{\tt t = 10 \; seconds}}}}$

∴ The stones will meet after time (t) 10 seconds.

$\rule{300}{1.5}$