Question

Two blocks each of masses 100g and 250g having the same size are dropped simultaneously from 81m and

100m respectively. Find the ratio of time they would take in reaching the ground. Will the ratio remain same

if (i) one of the objects is hollow and other one is solid and (ii) both of them are hollow, size remaining the

same in each case? Give reason.l​

Answer 1

Answer:

it is the question of kinematics it means it dont with the cause of the motion and it only depends on the height therefore

81 = 1/2 x 10 x t^2

81/5 = t^2

4.02 sec = t1

for the other block the displacement is 100

then

100 = 1/2 x 10 x t^2

100/5 = t^2

4.4 sec = t2

Answer 2

Solution :-

As per the given data ,

• Mass of block 1 ( m 1 ) = 100 g
• Height of mass 1 from the ground ( h1 ) = 81 m
• Mass of block 2 ( m 2 ) = 250 g

• Height of mass 2 from the ground ( h2 ) = 100 m
• Initial velocity of the block ( u) = 0 m/s ( dropped )

As the block is moving with uniform acceleration throughout it's motion we can apply the second equation of motion in order to find time

Hence ,

$\dag\boxed{\mathtt{h = ut + \dfrac{1}{2}gt^{2}}}$

As the initial velocity of the ball is zero ,

$\longrightarrow{\mathtt{h_1 = \dfrac{1}{2}gt_1^{2}}}...(1)$

Similarly ,

$\longrightarrow{\mathtt{h_2= \dfrac{1}{2}gt_2^{2}}}...(2)$

Dividing ( 1) and (2) we get ,

$\longrightarrow{\mathtt{\dfrac{h_1}{h_2} = (\dfrac{t_1}{t_2})^{2}}}$

$\longrightarrow\mathtt{\dfrac{t_1}{t_2} = \sqrt{\dfrac{h_1}{h_2}}}$

Now let's substitute the given values in the above equation ,

$\longrightarrow\mathtt{\dfrac{t_1}{t_2} = \sqrt{\dfrac{81}{100}}}$

$\longrightarrow\boxed{\red{\mathtt{\dfrac{t_1}{t_2} = \dfrac{9}{10}}}}$

From this we can conclude that , time depends only upon height

Hence , the ratio of time taken will remain the same in both the cases