A balloon is at a height of $81 ext{ m}$ and is ascending upwards with a velocity of $12 ext{ m/s}$. A bod

Question

A balloon is at a height of $81 	ext{ m}$ and is ascending upwards with a velocity of $12 	ext{ m/s}$. A body of $2 	ext{ kg}$ weight is dropped from it. If $g=10 	ext{ m/s}^2$, the body will reach the surface of the earth in:

Accepted Answer

1. **Identify Initial Conditions:** When an object is dropped from a moving body (like a balloon), it acquires the initial velocity of that body at the instant of release. 
   - Initial velocity ($u$) = $+12 	ext{ m/s}$ (directed upwards).
   - Displacement ($s$) = $-81 	ext{ m}$ (taking upward direction as positive, the final position (ground) is $81 	ext{ m}$ below the point of release).
   - Acceleration ($a$) = $-g = -10 	ext{ m/s}^2$ (acting vertically downwards).
   - The mass of the body ($2 	ext{ kg}$) does not affect the kinematics of free fall (neglecting air resistance) .
2. **Apply Equation of Motion:** Use the displacement-time relation: $s = ut + \frac{1}{2}at^2$ .
   $$ -81 = 12t + \frac{1}{2}(-10)t^2 $$
   $$ -81 = 12t - 5t^2 $$
3. **Solve the Quadratic Equation:** Rearrange the terms to form $5t^2 - 12t - 81 = 0$. Using the quadratic formula:
   $$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
   $$ t = \frac{12 \pm \sqrt{(-12)^2 - 4(5)(-81)}}{2(5)} $$
   $$ t = \frac{12 \pm \sqrt{144 + 1620}}{10} = \frac{12 \pm \sqrt{1764}}{10} $$
   $$ t = \frac{12 \pm 42}{10} $$
   Possible values: $t_1 = 5.4 	ext{ s}$ and $t_2 = -3 	ext{ s}$. Since time cannot be negative, $t = 5.4 	ext{ s}$.

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