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.

Given:

$v_x = 10$

m/s,

$g = 10$

m/s²,

$t = 1$

s

Vertical velocity:

$v_y = g t = 10$

m/s

Resultant velocity:

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 10^2} = 14.1 \text{ m/s}$$

• Time to fall:
  $t = \sqrt{2h/g} = \sqrt{90/10} = 3$ 

  s

• Vertical velocity:
  $v_y = gt = 10 \times 3 = 30$ 

  m/s

• Resultant velocity:
  $v = \sqrt{v_x^2 + v_y^2} = \sqrt{40^2 + 30^2} = 50$ 

  m/s

Answer: 50 m/s

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).

For meeting, equate positions: \(X_1 = X_2  -3 + 2t + t^2 = 7 - 8t + t^2;  10t = 10; t = 1 \text{ s} \). Differentiating positions gives velocities: \(v_1 = 2+2t = 4\text{ m/s}\) and \(v_2 = -8+2t = -6\text{ m/s}\) at \(t=1\text{ s}\). Relative speed \(V = |v_1 - v_2| = 10 \text {m/s}\).

Differentiating \(t = ax^2 + bx\) with respect to \(t\) gives \(1 = (2ax + b)v\). Differentiating again with respect to \(t\) gives \(a = \frac{dv}{dt} = -2av^3\).

We have \(v^2 = 180 - 16x\). Comparing with \(v^2 = u^2 + 2ax\), we get \(2a = -16 ⇒ a = -8\text{ ms}^{-2}\).

Initial velocity \(u = 72\text{ km/h} = 20\text{ m/s}\). Using \(s = ut - \frac{1}{2}at^2\): \(20 = 20(2) - \frac{1}{2}a(2)^2 ⇒ 20 = 40 - 2a ⇒a = 10\text{ ms}^{-2}\).