A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food

Q: Why does the packet have an initial upward velocity?

Because it was inside the helicopter, sharing its state of motion. When released, inertia keeps it moving upwards initially before gravity pulls it down.

Q: Why is the displacement taken as $-h$?

If we take the release point as the origin $(0,0)$ and upward as positive, the ground is located at a coordinate of $-h$. The net displacement is independent of the path taken.

Q: Could we solve this by splitting the motion into two parts (going up, then coming down)?

Yes, you can find the time to reach the peak and then the time to fall from the peak to the ground. But using the vector equation $s = ut + \frac{1}{2}at^2$ handles the entire journey elegantly in one single step!

Q: What does the negative root of the quadratic equation represent?

It represents the time in the past when the packet would have been at the ground if it had been following this exact parabolic trajectory from the very beginning.

Q: Why did we use $g$ for the helicopter's acceleration?

The problem explicitly states the helicopter rises with a constant acceleration $g$. This is just a given value for the engine's thrust, not to be confused with the acceleration due to gravity acting downwards.

Question

A helicopter rises from rest on the ground vertically upwards with a constant acceleration $g$. A food packet is dropped from the helicopter when it is at a height $h$. The time taken by the packet to reach the ground is close to (Here, $g$ is the acceleration due to gravity).

Options:

(A) $t = \frac{2}{3}\sqrt{\frac{h}{g}}$

(B) $t = 1.8\sqrt{\frac{h}{g}}$

(C) $t = \sqrt{\frac{2h}{3g}}$

(D) $t = 3.4\sqrt{\frac{h}{g}}$

Sigma Percentile · Accepted Answer

### The Setup: A Rising Helicopter

Imagine a helicopter starting from rest on the ground and accelerating vertically upwards.

The problem states that the helicopter has a constant upward acceleration of $g$.

We need to find out exactly how fast it is moving when it reaches a height $h$.

To do this, we can use the third equation of motion:

$$v^2 = u^2 + 2as$$

Since the helicopter starts from rest, its initial velocity $u = 0$.

Substituting the acceleration $a = g$ and displacement $s = h$, we get:

$$v^2 = 0 + 2gh$$

Taking the square root, the velocity of the helicopter at height $h$ is:

$$v = \sqrt{2gh}$$

### The Drop: Inertia in Action

Now comes the most critical part of the problem.

When the food packet is dropped from the helicopter, it does not simply fall from rest.

Because the packet was inside the moving helicopter, it shares the helicopter's state of motion.

**Due to the inertia of motion, the packet inherits the helicopter's upward velocity.**

Therefore, the initial velocity of the packet the moment it is released is $u = \sqrt{2gh}$ directed upwards.

### The Free Fall: Setting up the Math

Once released, the packet is in free fall under the influence of gravity.

It will travel upwards for a brief moment, reach a maximum height, and then fall all the way down to the ground.

Let's set up our sign convention. We will take the upward direction as positive.

The packet is released at height $h$ and lands on the ground, so its net displacement is downwards.

**Thus, the displacement is $s = -h$.**

The acceleration acting on the packet is due to gravity, which is downwards.

**So, the acceleration is $a = -g$.**

We can now use the second equation of motion to find the time $t$:

$$s = ut + \frac{1}{2}at^2$$

Substituting our values into the equation:

$$-h = (\sqrt{2gh})t - \frac{1}{2}gt^2$$

### The Resolution: Solving for Time

Let's rearrange this equation into a standard quadratic form $At^2 + Bt + C = 0$:

$$\frac{1}{2}gt^2 - \sqrt{2gh}t - h = 0$$

This might look intimidating, but we can solve it easily using the quadratic formula:

$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Plugging in our coefficients $a = \frac{g}{2}$, $b = -\sqrt{2gh}$, and $c = -h$:

$$t = \frac{\sqrt{2gh} \pm \sqrt{2gh - 4(\frac{g}{2})(-h)}}{2(\frac{g}{2})}$$

Notice how beautifully the terms inside the square root simplify.

The negative signs cancel out, giving us:

$$t = \frac{\sqrt{2gh} \pm \sqrt{2gh + 2gh}}{g}$$

$$t = \frac{\sqrt{2gh} \pm \sqrt{4gh}}{g}$$

We can pull out a factor of $\sqrt{gh}$ from the numerator:

$$t = \frac{\sqrt{gh}(\sqrt{2} \pm 2)}{g}$$

Simplifying the $g$ terms, we get:

$$t = \sqrt{\frac{h}{g}}(\sqrt{2} \pm 2)$$

Since time cannot be negative, we must reject the negative root $(\sqrt{2} - 2)$.

We take the positive root:

$$t = \sqrt{\frac{h}{g}}(\sqrt{2} + 2)$$

We know that $\sqrt{2} \approx 1.414$.

Adding $2$ to this gives $3.414$.

**Therefore, the total time taken by the packet to reach the ground is approximately $3.4 \sqrt{\frac{h}{g}}$.**

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Velocity of Helicopter$

$Calculating v$

$Initial Velocity of Packet$

$Equation of Motion for Packet$

$Substituting Values$

$Forming Quadratic Equation$

$Applying Quadratic Formula$

$Simplifying Roots$

$Extracting g h$

$Final Calculation$

The Sigma Insight: Motion in a Straight Line

The Setup

The Drop

The Free Fall

The Resolution

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of Helicopter

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Velocity of Packet

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Motion for Packet

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forming Quadratic Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Quadratic Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Roots

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extracting gh​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation