A person in an elevator accelerating upwards with an acceleration of $2$ m s$^{-2}$, tosses a coin vertically upwards with a speed of 20 m s$^{1}$. After how much time will the coin fall back into his hand? ($g = 10$ m s$^{-2}$)

A person in an elevator accelerating upwards with an acceleration of 2 m s⁻², tosses a coin vertically upwards

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Concept understanding — Projectile Motion Under Gravity

Projectile Motion Under Gravity

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it curves upward, then arcs downward. That curve is a parabola, and the motion is called projectile motion.

The key insight: once the ball leaves your hand, the only force acting on it (ignoring air resistance) is gravity pulling it straight down. There is no force pushing it sideways or upward after release. That single downward force is what creates the beautiful curved path.

The Core Idea

A projectile is any object that is thrown, launched, or otherwise projected into the air and then moves under the influence of gravity alone. The motion has two independent parts happening simultaneously:

Horizontal motion: constant speed (no horizontal force)
Vertical motion: constant downward acceleration $g \approx 9.8 \, \text{m/s}^2$

These two motions are completely independent — they don't affect each other. This is the most important thing to understand.

Important

The horizontal and vertical motions are independent. The horizontal speed stays constant; the vertical speed changes by $9.8 \, \text{m/s}$ every second downward.

Breaking It Down Mathematically

Let's set up coordinates: $x$ is horizontal, $y$ is vertical (positive upward). The launch point is at $(0,0)$ with initial speed $u$ at angle $\theta$ above horizontal.

Initial velocity components:

u_x = u \cos \theta

u_y = u \sin \theta

Horizontal motion (no acceleration):

x = u_x t = (u \cos \theta) t

Vertical motion (constant downward acceleration $g$ ):

y = u_y t - \frac{1}{2} g t^2 = (u \sin \theta) t - \frac{1}{2} g t^2

The minus sign is because gravity pulls downward, opposite to our positive $y$ direction.

The Path Is a Parabola

Eliminate $t$ between the $x$ and $y$ equations. From $x = u_x t$ , we get $t = \frac{x}{u \cos \theta}$ . Substitute into the $y$ equation:

y = (u \sin \theta) \left( \frac{x}{u \cos \theta} \right) - \frac{1}{2} g \left( \frac{x}{u \cos \theta} \right)^2

y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2

This is of the form $y = ax - bx^2$ , which is a parabola opening downward. That's why every projectile under gravity follows a parabolic path.

y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2

Key Quantities You'll Need

Time of Flight ( $T$ )

The total time the projectile stays in the air. Set $y = 0$ (returns to launch height):

0 = (u \sin \theta) T - \frac{1}{2} g T^2

Factor $T$ : $T (u \sin \theta - \frac{1}{2} g T) = 0$

The non-zero solution:

T = \frac{2 u \sin \theta}{g}

Maximum Height ( $H$ )

The highest point occurs when vertical velocity becomes zero: $v_y = u \sin \theta - g t = 0$ , so $t = \frac{u \sin \theta}{g}$ .

Plug into $y$ equation:

H = (u \sin \theta) \left( \frac{u \sin \theta}{g} \right) - \frac{1}{2} g \left( \frac{u \sin \theta}{g} \right)^2

H = \frac{u^2 \sin^2 \theta}{g} - \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin^2 \theta}{2g}

Range ( $R$ )

Horizontal distance traveled when it returns to launch height. Use $x = u_x T$ :

R = (u \cos \theta) \cdot \frac{2 u \sin \theta}{g} = \frac{2 u^2 \sin \theta \cos \theta}{g}

Using $\sin 2\theta = 2 \sin \theta \cos \theta$ :

R = \frac{u^2 \sin 2\theta}{g}

Tip

Maximum range occurs when $\sin 2\theta = 1$ , i.e., $2\theta = 90^\circ$ or $\theta = 45^\circ$ . At this angle, $R_{\text{max}} = \frac{u^2}{g}$ .

Common Mistakes to Avoid

Watch out

Don't mix up horizontal and vertical equations. Horizontal has constant speed; vertical has constant acceleration. …

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The coin’s motion is analyzed relative to the accelerating elevator using the concept of relative acceleration. The time for the coin to return to the hand is $\frac{10}{3}$ seconds.

Concept and Intuition

This problem is about projectile motion, but with a twist: the observer (the person in the elevator) is accelerating upward. When you toss a coin upward inside an accelerating elevator, the coin’s motion relative to the ground is governed by gravity alone (downward acceleration $g = 10$ m/s²). However, the elevator floor (and the person’s hand) is accelerating upward at $2$ m/s². So, relative to the person, the coin experiences an effective downward acceleration of $g + a = 12$ m/s². This is because the hand is “chasing” the coin upward, making the coin appear to fall back faster than it would in a stationary frame.

The key insight: treat the problem in the non-inertial frame of the elevator, where the coin’s relative acceleration is $g + a$ downward. The initial relative velocity is the toss speed upward (20 m/s). The coin returns to the hand when its relative displacement becomes zero again.

Step-by-Step Solution

Choose the reference frame.

Work in the frame of the elevator (attached to the person’s hand). In this frame, the hand is at rest. The coin is tossed upward with initial speed $u = 20$ m/s relative to the hand.
Determine the effective acceleration of the coin in this frame.

The elevator accelerates upward at $a = 2$ m/s² relative to the ground. Gravity pulls the coin downward at $g = 10$ m/s² relative to the ground. In the elevator’s frame, the coin’s acceleration is the sum of these two downward accelerations (since the frame itself accelerates upward, it adds an extra downward pseudo-force).

So, relative acceleration $a_{\text{rel}} = g + a = 10 + 2 = 12$ m/s² downward.
Set up the equation of motion.

Let upward direction be positive. Initial relative velocity $u = +20$ m/s. Relative acceleration $a_{\text{rel}} = -12$ m/s² (negative because downward). The relative displacement $s$ of the coin from the hand at time $t$ is given by:

$s = u t + \frac{1}{2} a_{\text{rel}} t^2$ …

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Q.A person in an elevator accelerating upwards with an acceleration of $2$ m s $^{-2}$ , tosses a coin vertically upwards with a speed of 20 m s $^{1}$ . After how much time will the coin fall back into his hand? ( $g = 10$ m s $^{-2}$ )

Concept understanding — Projectile Motion Under Gravity

Projectile Motion Under Gravity

The Core Idea

Breaking It Down Mathematically

The Path Is a Parabola

Key Quantities You'll Need

Time of Flight ( $T$ )

Maximum Height ( $H$ )

Range ( $R$ )

Common Mistakes to Avoid

Concept and Intuition

Step-by-Step Solution

Concept: Motion in a Non-Inertial (Accelerating) Frame

Q.A person in an elevator accelerating upwards with an acceleration of 222 m s−2^{-2}−2, tosses a coin vertically upwards with a speed of 20 m s1^{1}1. After how much time will the coin fall back into his hand? (g=10g = 10g=10 m s−2^{-2}−2)

Concept and Intuition

Step-by-Step Solution

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Q.A person in an elevator accelerating upwards with an acceleration of $2$ m s $^{-2}$ , tosses a coin vertically upwards with a speed of 20 m s $^{1}$ . After how much time will the coin fall back into his hand? ( $g = 10$ m s $^{-2}$ )

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