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:

1. $u = \sqrt{2gH} \Rightarrow u = \sqrt{gH}$ 
2. $-H = ut - \frac{1}{2}gt^2 \Rightarrow gt^2 - 2ut - 2H = 0$
3. Substituting $u = \sqrt{gH}$ into the quadratic: 
   $gt^2 - 2\sqrt{gH}t - 2H = 0$
4. Rewriting: $gt^2 - 2\sqrt{gH}t + \sqrt{gH}t - 2H = 0$
5. Grouping terms: $\sqrt{g}t(\sqrt{g}t - 2\sqrt{H}) + \sqrt{H}(\sqrt{g}t - 2\sqrt{H}) = 0$
6. Factoring: $(\sqrt{g}t + \sqrt{H})(\sqrt{g}t - 2\sqrt{H}) = 0$
7. So, $\sqrt{g}t = 2\sqrt{H} \Rightarrow t = \frac{2\sqrt{H}}{\sqrt{g}}$

Now use this relation:
Let $t = \frac{5}{2}$ sec and $g = 32$ ft/s² (standard gravity). Then, 
$t = \frac{2\sqrt{H}}{\sqrt{g}} \Rightarrow \frac{5}{2} = \frac{2\sqrt{H}}{\sqrt{32}}$

Solving for H:
$\frac{5}{2} = \frac{2\sqrt{H}}{4\sqrt{2}} \Rightarrow \frac{5}{2} = \frac{\sqrt{H}}{2\sqrt{2}}$
$\Rightarrow \sqrt{H} = 5\sqrt{2} \Rightarrow H = (5\sqrt{2})^2 = 25 \cdot 2 = \mathbf{50}$ ft

Correction: There's a mistake in previous options if $H$ is 50 ft, but for $H = 112.5$ ft and $t = \frac{5}{2}$ sec to be correct:

Use $t = \frac{2\sqrt{H}}{\sqrt{g}} \Rightarrow \frac{5}{2} = \frac{2\sqrt{H}}{\sqrt{32}}$ 
$\Rightarrow \frac{5}{2} = \frac{2\sqrt{H}}{4\sqrt{2}} \Rightarrow \sqrt{H} = \frac{5}{2} \cdot 2\sqrt{2} = 5\sqrt{2}$ 
$\Rightarrow H = (5\sqrt{2})^2 = \mathbf{50}$ ft, not 112.5 ft.

Therefore, only option:
(C): $T = \frac{5}{2}$ sec is correct.
If the original answer says (B): $H = 112.5$, then either $g$ or time value assumed is different.

Answer 2

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:
$s = ut + \frac{1}{2}at^2$
$s = 0 \times 5 + \frac{1}{2} \times 4 \times 5^2 = 2 \times 25 = \mathbf{50 \text{ ft}}$

Step 2: Calculate the velocity of the balloon after 5 seconds
$v = u + at = 0 + 4 \times 5 = \mathbf{20 \text{ ft/sec}}$ upward

Step 3: Time taken for the stone to reach the ground
The stone is dropped from a height of 50 ft with $g = 32$ ft/sec².
$s = ut + \frac{1}{2}gt^2$ with $u = 0$:
$50 = 16t^2 \Rightarrow t^2 = \frac{25}{8} \Rightarrow t = \frac{5}{2}$ sec
So, the stone takes $\mathbf{\frac{5}{2} \text{ sec}}$ to reach the ground.

Step 4: Additional height balloon ascends during stone's fall
$s = ut + \frac{1}{2}at^2 = 20 \times \frac{5}{2} + \frac{1}{2} \times 4 \times \left(\frac{5}{2}\right)^2$
$s = 50 + 2 \times \frac{25}{4} = 50 + 12.5 = \mathbf{62.5 \text{ ft}}$

Step 5: Total height of the balloon when the stone hits the ground
$H = 50 + 62.5 = \mathbf{112.5 \text{ ft}}$

Final Answers:
- (B): $\mathbf{H = 112.5}$ ft
- (C): $\mathbf{T = \frac{5}{2}}$ sec