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 time taken to fall through height \(h\) from rest under gravity is \(t = \sqrt{\frac{2h}{g}}\). Since \(t \propto \sqrt{h}\), the ratio of times for heights \(a\) and \(b\) is \(\sqrt{a} : \sqrt{b}\).

Horizontal range is given by \(R = \frac{u^2 \sin(2\theta)}{g}\). For complementary projection angles \(\theta\) and \(90^\circ - \theta\) (like \(30^\circ\) and \(60^\circ\)), the horizontal ranges are equal. Thus, the ratio of range is \(1 : 1\).