Question

25. An object is dropped from rest at a height of 150 m and simultaneously another
object is dropped from rest at a height 100 m. What is the difference in their heights
after 2 s if both the objects drop with same accelerations? How does the difference
in heights vary with time?​

Answer 1

Answer:

The difference in heights is 50m.

Given:

u = 0

t = 2 sec

To find:

How does the difference in heights vary with time ?

Solution:

We know that,

From second equation of motion, distance covered by first object in 2 sec is :-

$\qquad{\underline{\boxed{\sf h = ut + \dfrac{1}{2} gt^2}}}$

$\quad{\pink{\sf Take \ g = 10 \ m/s^2}}$

$\quad \longrightarrow \sf h = 0 + \dfrac{1}{2} \times 10 \times 4$

Height of first object from ground after 2 sec ($\sf h_1$) = 150m - 20m = 130m.

For second object, u = 0 and t = 2 sec.

From second equation of motion,

$\qquad {\underline{\boxed{\sf h = ut + \dfrac{1}{2} gt^2}}}$

$\quad \longrightarrow \sf h = 0 \times 2 + \dfrac{1}{2} \times 10 \times (2)^2$

$\quad \longrightarrow \sf h = 0 + \dfrac{1}{2} \times 10 \times 4$

$\quad \therefore {\underline{\boxed{\sf h = 20m}}}$

Height of second object from ground after 2 sec,

$\quad \sf h_2 = 100m - 20m$

$\quad \sf h_2 = 80m$

Difference in height after two sec,

$\qquad \longrightarrow \sf h_1 - h_2$

$\qquad \longrightarrow 130-80$

$\qquad \therefore {\underline{\boxed{\sf h = 50m}}}$

$\therefore$ Distance in height vary with time is 50 meter.

Answer 2

Answer:

The difference in heights is 50m.

Given:

u = 0

t = 2 sec

To find:

How does the difference in heights vary with time ?

Solution:

We know that,

From second equation of motion, distance covered by first object in 2 sec is :-

$\qquad{\underline{\boxed{\sf h = ut + \dfrac{1}{2} gt^2}}}$

$\quad{\pink{\sf Take \ g = 10 \ m/s^2}}$

$\quad \longrightarrow \sf h = 0 + \dfrac{1}{2} \times 10 \times 4$

Height of first object from ground after 2 sec ($\sf h_1$) = 150m - 20m = 130m.

For second object, u = 0 and t = 2 sec.

From second equation of motion,

$\qquad {\underline{\boxed{\sf h = ut + \dfrac{1}{2} gt^2}}}$

$\quad \longrightarrow \sf h = 0 \times 2 + \dfrac{1}{2} \times 10 \times (2)^2$

$\quad \longrightarrow \sf h = 0 + \dfrac{1}{2} \times 10 \times 4$

$\quad \therefore {\underline{\boxed{\sf h = 20m}}}$

Height of second object from ground after 2 sec,

$\quad \sf h_2 = 100m - 20m$

$\quad \sf h_2 = 80m$

Difference in height after two sec,

$\qquad \longrightarrow \sf h_1 - h_2$

$\qquad \longrightarrow 130-80$

$\qquad \therefore {\underline{\boxed{\sf h = 50m}}}$

$\therefore$ Distance in height vary with time is 50 meter.