Question

Two objects of masses m1 and m2 having the same size are dropped simultaneously from heights h1 and h2 respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if (i) one of the objects is hollow and the other one is solid and (ii) both of them are hollow, size remaining the same in each case. Give reason.

Answer 1

h1 = 12 gt21

h2 = 12gt22

as x = o

t1t2 2 = √h1/h2

As the acceleration remains the same, the ratio between two objects remains the same. In this case, acceleration does not depend upon mass and size.

PLZ MARK ME AS BRAINLIST DUDE

ITS URGENT

Answer 2

Solution :-

As per the given data ,

• mass of object 1 = m 1

• mass of object 2 = m 2

• initial velocity for both the cases = 0 m/s (as the ball is dropped )

As the motion of both the masses is uniform we can use the second kinematic equation in order to solve this question,

$\bigstar\boxed{\mathtt{ h = ut + \dfrac{1}{2} g {t}^{2} }}$

here ,

• h = height

• u = initial velocity

• t = time

• g = acceleration due to gravity

Part I :-

$\longrightarrow\mathtt{h_1 = ut_1 + \dfrac{1}{2} g {t_1}^{2} }$

$\longrightarrow\mathtt{h_1 = 0\times t_1 + \dfrac{1}{2} g {t_1}^{2} }$

$\longrightarrow\mathtt{h_1 = \dfrac{1}{2} g {t}^{2} }$

$\longrightarrow\mathtt{t_1 = \dfrac{2h_1}{g} }$

Part II :-

Similarly as Part I we get ,

$\longrightarrow\mathtt{t_2= \dfrac{2h_2}{g} }$

From both these equations we can conclude that time taken only depends upon the height Hence , the time taken for both the masses will be same in both the cases