Question

A stone is allowed to fall from a tower of height

200 m and at the same time another stone is

projected vertically upwards from the ground at

a velocity of 20 m/s. Calculate when and where

the stones will meet.​

Answer 1

Answer:

Please follow the figure in attachment!

figure explanation:

height given is 200 m

let a be the point in which the 2 balls will meet in the future !!

so,one part be x..so,other part will be 200 -x

so,ball on the upper part will have initial velocity as 0..

from 2nd eqn of motion,

×=ut + at^2

u=0 ..so, x= 1/2at^2

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

x=1/2 ×10× t^2

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

for lower stone,

u=20

so, x=200-x

from 2nd eqn of motion

x=ut +1/2at^2

200-x= 20× t -1/2gt^2 (a=-g)

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

sub (1 ) in 2::

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

t =10 sec

sub t in 1

x= 5× 10×10

x=500m

therefore, the 2 stones will meet in 500m at 10th sec

Answer 2

The stones will meet at 77.5 m height in 5 sec.

Explanation:

Given that,

Height = 200 m

Suppose the velocity is 40 m/s

Velocity = 40 m/s

Let the two stones meet at time t at a height h above the ground.

We need to calculate the time

Using equation of motion

For stone A,

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$(200-h)=0+\dfrac{1}{2}\times9.8 t^2$

$200-h=4.9t^2$....(I)

For stone B,

Using equation of motion again

$s=ut-\dfrac{1}{2}gt^2$

Put the value in the equation

$h=40t-4.9t^2$....(II)

Adding equation (I) and (II)

$200-h+h=4.9t^2+40-4.9t^2$

$t=5\ sec$

Now, put the value of t in equation (II)

$h=40\times5-4.9\times(5)^2$

$h=77.5\ m$

Hence, The stones will meet at 77.5 m height in 5 sec.

Learn more :

Topic : kinematics

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