Using the stopping distance formula:

$$s \propto u^2$$

If speed is halved, new stopping distance:

$$s_2 = \frac{100}{4} = 25 \text{ m}$$

Using formula for Time:

$$t = \sqrt{\frac{2h}{g}}$$

t=g2h​​

$$t = \sqrt{\frac{2 \times 80}{10}} = \sqrt{16} = 4 \text{ s}$$

t=102×80​​=16​=4 s

Correct Answer: (1) 4 s

Using the formula for range in horizontal projectile motion:

$$R = v_x \times \sqrt{\frac{2h}{g}}$$

R=vx​×g2h​​

$$R = 20 \times \sqrt{\frac{2 \times 80}{10}}$$

R=20×102×80​​

$$R = 20 \times \sqrt{16} = 20 \times 4 = 80 \text{ m}$$

R=20×16​=20×4=80 m

Correct Answer: (3) 80 m

Using the direct range formula for a projectile dropped from a height:

$$R = v_x \times \sqrt{\frac{2h}{g}}$$

$$R = 10 \times \sqrt{\frac{2 \times 125}{10}}$$

​​

$$R = 10 \times \sqrt{25} = 10 \times 5 = 50 \text{ m}$$

\[ y = -\frac{g}{2 v_x^2} x^2 \]

This is of the form

$y = ax^2$

, which represents a parabolic path. Hence, the object follows a parabolic trajectory when viewed from the ground.

From the helicopter’s frame of reference, both the helicopter and the package have the same horizontal velocity. Since the package moves downward only due to gravity, it appears to fall straight down in a vertical line when observed from the helicopter.

The volume of a spherical bubble is \(V = \frac{4}{3}\pi r^3\). Differentiating with respect to time gives \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\). Given \(r = 0.5t\), we have \(\frac{dr}{dt} = 0.5\), and at \(t = 4\text{ s}\), \(r = 2\). Thus, \(\frac{dV}{dt} = 4\pi (2)^2 (0.5) = 8\pi\text{ units/s}\).

Distance in \( n^{\text{th}} \) second is \( S_n = u + frac {a}{2}(2n-1)) \). Starting from rest \( u=0 , S_n =frac {a}{2}(2n-1).\) Given

\( S_n = 3 S_{n-1} \), we have 2n-1 = 3(2n-3). Solving gives (4n = 8), hence (n = 2).

Stopping distance \(s\) is proportional to \(u^2\). When the speed is doubled (from 50 to 100), the stopping distance increases by a factor of 4. Thus, \(s' = 4 \times 6 = 24\text{ m}\).