Question

A balloon starting from rest ascends from the ground with a uniform acceleration of 4 ft/sec². At the end of 5 sec, a stone is dropped from it. If t be the time to reach the stone to the ground and H is the height of the balloon when the stone goes to the floor, Then

• A. T=6 sec
• B. H=112.5 ft.
• C. T=\(\frac{5}{2}\) sec
• D. 225 ft.

Answer 1

The Correct Option is B, C

Given the equations:

1. u=2​×gH​⇒u=gH​
2. −H=ut−21​gt2⇒gt2−2ut−2H=0
3. gt2−gH​t2−2H=0
4. gt2−2gH​t+gH​t−2H=0
5. gt​[gt​−2H​]+H​[gt​−2H​]=0
6. (gt​+H​)(gt​−2H​)=0
7. t=2√H/g​​

The correct answer is/are option(s):
(B): H=112.5 ft.
(C):  T=\(\frac{5}{2}\) sec

Answer 2

Solution
 Step 1: Calculate the height of the balloon after 5 seconds
- Initial velocity (\( u \)) = 0 ft/sec (starting from rest)
- Acceleration (\( a \)) = 4 ft/sec²
- Time (\( t \)) = 5 sec
Using the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Substitute the given values:
\[ s = 0 \times 5 + \frac{1}{2} \times 4 \times 5^2 \]
\[ s = 0 + \frac{1}{2} \times 4 \times 25 \]
\[ s = 2 \times 25 \]
\[ s = 50 \text{ ft} \]
So, the height of the balloon after 5 seconds is 50 ft.
Step 2: Calculate the velocity of the balloon after 5 seconds
Using the equation:
\[ v = u + at \]
Substitute the given values:
\[ v = 0 + 4 \times 5 \]
\[ v = 20 \text{ ft/sec} \]
So, the velocity of the balloon after 5 seconds is 20 ft/sec upward.
Step 3: Calculate the time taken for the stone to reach the ground
The stone is dropped from a height of 50 ft and falls freely under gravity. The acceleration due to gravity (\( g \)) is 32 ft/sec².
Using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
Since the initial velocity of the stone (\( u \)) is 0 (it is dropped), the equation simplifies to:
\[ 50 = 0 \times t + \frac{1}{2} \times 32 \times t^2 \]
\[ 50 = 16t^2 \]
\[ t^2 = \frac{50}{16} \]
\[ t^2 = \frac{25}{8} \]
\[ t = \sqrt{\frac{25}{8}} \]
\[ t = \frac{5}{2} \text{ sec} \]
So, the stone takes \( \frac{5}{2} \) seconds to reach the ground.
 Step 4: Calculate the additional height the balloon ascends during the stone's fall
During the \( \frac{5}{2} \) seconds that the stone is falling, the balloon continues to ascend with an initial velocity of 20 ft/sec and an acceleration of 4 ft/sec².
Using the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Substitute the given values:
\[ s = 20 \times \frac{5}{2} + \frac{1}{2} \times 4 \times \left(\frac{5}{2}\right)^2 \]
\[ s = 20 \times 2.5 + \frac{1}{2} \times 4 \times \frac{25}{4} \]
\[ s = 50 + 2 \times 6.25 \]
\[ s = 50 + 12.5 \]
\[ s = 62.5 \text{ ft} \]
So, the additional height ascended by the balloon during the stone's fall is 62.5 ft.
Step 5: Calculate the total height when the stone reaches the ground
The total height \( H \) is the sum of the height of the balloon after 5 seconds and the additional height ascended during the stone's fall:
\[ H = 50 \text{ ft} + 62.5 \text{ ft} \]
\[ H = 112.5 \text{ ft} \]
Final Answers is Option B and C
- Time taken for the stone to reach the ground (\( t \)) = \( \frac{5}{2} \) sec
- Height of the balloon when the stone reaches the ground (\( H \)) = 112.5 ft