Question

An object is dropped from rest at a height of 160m and simultaneously another object is dropped from rest at a height of 100m. what is the differences in their height after 2sce.how does the difference in height vary with time ​

Answer 1

Answer:

For the 1st object, u1=0, t1=2sec

The object is 160m above the ground.

h1=160m

a=g=10 (9.8)

$s1 = u1t1 + \frac{1}{2} a {t1}^{2}$

s1=(0×2 + 1/2×10×2×2)m

s1=(0+20)m

s1=20m

After 10 sec the distance between the object and ground is ,d1 = h1-s1

=(160-20)m=140m

Similarly, for the 2nd object , u2=0,t2= 2sec

a=g=10

The object is 100m above the ground.

h2=100m

$s2 = u2t2 + \frac{1}{2}a {t2}^{2}$

s2= (0×2 + 1/2×10×2×2)m

s2=(0+20)m

s2= 20m

After 2sec the distance between the object and ground is , d2=h2-s2

= (100-20)=80m

The distance between the object after 2sec is , d1-d2=(140-80)m= 50m

So the difference is same i.e. 50m

Hence we can concluded that the difference in height of the 2 objects does not depend on time and will always be same.

Hope it helps! ! !