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Question #42068

Difficulty: 8/10 | 🟢 Free

The Question

An elevator is accelerating with an acceleration $\vec{a} = 5\sqrt{3} \hat{i} + 5 \hat{j} \text{ m/s}^2$ , where $\hat{i}$ is along the horizontal floor of the elevator and $\hat{j}$ is vertically upwards. Inside this elevator, a projectile is fired from the origin on the floor with an initial speed of $20 \text{ m/s}$ at an angle of $60^\circ$ with the horizontal floor. Taking $g = 10 \text{ m/s}^2$ in the $-\hat{j}$ direction, what is the horizontal range of the projectile on the floor of the elevator?

Options

A.
$0 \text{ m}$
B.
$20 \text{ m}$
C.
$\frac{40\sqrt{3}}{3} \text{ m}$
D.
$20\sqrt{3} \text{ m}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Relative Velocity in 2D|Apparent Weight in Accelerating Elevators|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the application of pseudo-forces in 2D non-inertial frames.

The lethal trap is blindly applying standard projectile formulas like $R = u^2\sin 2\theta / |\vec{g}_{eff}|$ (which yields 20 m) or ignoring the horizontal component of the pseudo-force (which yields $40\sqrt{3}/3$ m).

By setting the launch velocity exactly anti-parallel to the effective gravity vector, the true relative trajectory is a straight line (1D motion) in the elevator frame, yielding a zero range on the floor.

This strongly penalizes formula-dependent approaches and rewards derivation from first principles.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Relative Velocity in 2D
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free