Animated Solution for Physics - Kinematics: A model rocket fired from the ground ascends with a constant upward acceleration. A small bolt is dropped from the rocket $1.0\text{ s}$ after the firing and fuel of the rocket is finished $4.0\text{ s}$ after the bolt is dropped. Air-time of the bolt is $2.0\text{ s}$. Acceleration of free fall is $10{\text{ m/s}}^2$. Which of the following statements is/are correct?

Imagine standing on the launchpad, watching a model rocket tear into the sky. It's accelerating upwards, fighting gravity with its engine. But this isn't just a simple launch; it's a multi-stage physics puzzle. Exactly one second into the flight, a small bolt shakes loose and drops. Four seconds later, the rocket's engine sputters and dies. Our mission? To decode every phase of this journey.

A Rocket and a Bolt

Let's define our variables. The rocket starts from rest and accelerates upwards with a constant acceleration, let's call it $a$.

At $t=1.0\text{ s}$, the bolt drops. What is the state of the rocket at this exact moment?

Using the first equation of motion, its velocity is $v_1=a\cdot (1.0)=a$.

Its height above the ground is $y_1=\frac{1}{2}a(1.0{)}^2=0.5a$.

Here is the crucial physical insight: when the bolt drops, it doesn't just fall from rest. Due to inertia, it inherits the exact velocity of the rocket at that instant. So, the bolt begins its free fall with an initial upward velocity $u_b=a$ from a height of $0.5a$.

Decoding the Bolt's Journey

The bolt is now a projectile under the influence of gravity ($g=10{\text{ m/s}}^2$). We are told it stays in the air for $2.0\text{ s}$.

Let's apply the second equation of motion for the bolt:

Taking the upward direction as positive, the bolt's final displacement when it hits the ground is $-0.5a$.

Boom! We've unlocked the rocket's acceleration. Statement (a) is absolutely correct.

The Rocket's Powered Ascent

Now that we know $a=8.0{\text{ m/s}}^2$, we can analyze the rocket's powered flight. The fuel finishes $4.0\text{ s}$ after the bolt drops, which means the engine burned for a total of $t_2=1.0+4.0=5.0\text{ s}$.

How high did it go in this time?

Statement (b) is spot on. The fuel ran out exactly $100\text{ m}$ above the ground.

What about its speed? The rocket accelerates as long as the engine fires, so it reaches its maximum speed right at the $5.0\text{ s}$ mark.

Statement (c) is also correct!

The Free Fall to Earth

At $100\text{ m}$ in the air, traveling at $40\text{ m/s}$ upwards, the engine dies. The rocket is now in free fall. It will coast higher, reach a peak, and plummet back to Earth.

Let's find the time $t_3$ it takes to hit the ground from this point. We use the displacement equation again, with initial velocity $u=40\text{ m/s}$, displacement $s=-100\text{ m}$, and acceleration $a=-10{\text{ m/s}}^2$.

Dividing the entire equation by $-5$ to simplify:

This factors beautifully:

Since time cannot be negative, $t_3=10\text{ s}$.

The total air-time of the rocket is the powered time plus the free fall time:

Statement (d) is correct.

Every single statement provided in the question holds true. This problem is a masterclass in breaking down complex, multi-stage kinematics into manageable, logical steps. By carefully tracking the initial conditions of each phase, we unraveled the entire journey of both the rocket and the bolt.