Question

A ballon is ascending up with a velocity of 4 ms 1.An object is dropped from it when it is at a
height of 100 m above the ground. The distance between the object and ballon after 2 seconds is
(g=10 ms 2)
a) 10 m
b) 20 m
c) 30 m
d) 40 m
​

Answer 1

Given:

A ballon is ascending up with a velocity of 4 m/s, an object is dropped from it when it is at a height of 100 m above the ground.

To find:

Distance between object and balloon after 2 sec?

Calculation:

CONSIDER THE 100 m Line TO BE THE REFERENCE !!

Now, position of balloon after object is dropped:

$y1 =4 \times 2$

$\implies y1 = 8 \: m$

Now, position of object 2 sec after being dropped:

$y2 = ut - \dfrac{1}{2} g {t}^{2}$

$\implies y2 =(4 \times 2) - \dfrac{1}{2} g {(2)}^{2}$

$\implies y2 =8 - 20$

$\implies y2 = - 12 \: m$

Now, distance between them :

$y = |y1| + |y2|$

$\implies y = 8 + 12$

$\implies y = 20 \: m$

So, distance between them is 20 m.

Answer 2

Explanation:

Given:

A ballon is ascending up with a velocity of 4 m/s, an object is dropped from it when it is at a height of 100 m above the ground.

To find:

Distance between object and balloon after 2 sec?

Calculation:

CONSIDER THE 100 m Line TO BE THE REFERENCE !!

Now, position of balloon after object is dropped:

y1 =4 \times 2y1=4×2

\implies y1 = 8 \: m⟹y1=8m

Now, position of object 2 sec after being dropped:

y2 = ut - \dfrac{1}{2} g {t}^{2} y2=ut−

2

1

gt

2

\implies y2 =(4 \times 2) - \dfrac{1}{2} g {(2)}^{2} ⟹y2=(4×2)−

2

1

g(2)

2

\implies y2 =8 - 20⟹y2=8−20

\implies y2 = - 12 \: m⟹y2=−12m

Now, distance between them :

y = |y1| + |y2| y=∣y1∣+∣y2∣

\implies y = 8 + 12⟹y=8+12

\implies y = 20 \: m⟹y=20m

So, distance between them is 20 m.