Question

A bottle dropped from a balloon reaches the ground in 20 s. Determine the height of the balloon if (a) it was at rest in the air and (b) it was ascending with a speed of 50 m/s when the ball was dropped. Ans. (a) 2.0 km; (b) 0.96 km

Answer 1

We are given that ,

A bottle is dropped from a balloon which reached the ground in 20 seconds.

We have to find the height at which the balloon was present if

1. It was at rest in the air
2. It was ascending with of a speed of 50m/s

[1]

Here, In this case, we have

• Time taken to reach ground = 20 s
• Initial velocity = 0 m/s
• Acceleration due to gravity, g = 10 m/s²

Using the following formula, we get

⇒ T = v / g

⇒ 20 = v/10

⇒ v = 200 m/s

So, The final velocity of the bottle is 200 m/s and the initial velocity is 0.

Using the third equation of motion, we have

⇒ v² - u² = 2gs

⇒ 200² - 0 = 2×10×s

⇒ 40000 / 20 = s

⇒ s = 2000 m or 2 Km

[2]

Here, It is given that the balloon is ascending at 50 m/s hence the initial velocity of the bottle is -50 m/s

Total Time taken by the bottle to reach the ground = 20 seconds.

Using the second equation of motion,

⇒ s = ut + 1/2 gt²

⇒ s = ut + 1/2 gt²

⇒ s = -50 × 20 + 1/2 × 10 × 20²

⇒ s = - 1000 + 5 × 400

⇒ s = 2000 - 1000

⇒ s = 1000 m or 1 km.

Answer 2

$\sf{\pink{\underline{\underline{\purple{GIVEN:-}}}}}$

• A bottle dropped from a balloon reaches the ground in 20s .

$\sf{\pink{\underline{\underline{\purple{TO\:FIND:-}}}}}$

• Determine the height of the balloon, if

1. It was at rest in the air .
2. It was ascending with a speed of 50m/s when the ball was dropped .

$\sf{\pink{\underline{\underline{\purple{SOLUTION:-}}}}}$

☃️ In first case, given that it was at rest .

• initial velocity (u) = 0m/s

• Time (t) = 20s

• Acceleration (a) = acceleration due to gravity (g) = 10m/s²

$\orange\star\:\bf{\gray{\boxed{\boxed{\purple{S\:=\:ut\:+\:\dfrac{1}{2}\:at^2\:}}}}}$

Where,

• S = Distance or Height (H)

$\rm{\implies\:H\:=\:0\times{20}\:+\:\dfrac{1}{2}\times{10}\times{(20)^2}\:}$

$\rm{\implies\:H\:=\:0\:+\:5\times{400}\:}$

$\rm\purple{\implies\:Height\:=\:2000\:m\:}$

$\rm\pink{\therefore}$ [1] The height of the balloon, when it was at rest in air is “ 2000m ” .

☃️ In second case, given that balloon was ascending with a speed of 50m/s .

• initial velocity (u) = -50m/s

[NOTE :- -ve sign indicates that the balloon was dropped in downward direction which is towards the gravity .]

• Time (t) = 20s

$\orange\star\:\bf{\gray{\boxed{\boxed{\purple{S\:=\:ut\:+\:\dfrac{1}{2}\:at^2\:}}}}}$

Where,

• S = Distance or Height (H)

$\rm{\implies\:H\:=\:-50\times{20}\:+\:\dfrac{1}{2}\times{10}\times{(20)^2}\:}$

$\rm{\implies\:H\:=\:-1000\:+\:5\times{400}\:}$

$\rm{\implies\:H\:=\:-1000\:+\:2000\:}$

$\rm\purple{\implies\:H\:=\:1000\:m}$

$\rm\pink{\therefore}$ [2] The height of the balloon, when it was ascending with a speed of 50m/s is “ 1000m ”