JEE Main 2023 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 8 April 2023, Shift 2, Physics Q39

Question 39 of 90 in this shift, Physics question 9 of 30, Section A.

The trajectory of projectile, projected from the ground is given by

y = x - \frac{x^2}{20}

. Where

x

and

y

are measured in meter. The maximum height attained by the projectile will be.

(1)5 mOfficial answer
(2) $10\sqrt{2}$ m
(3)10 m
(4)200 m

Official answer

Option 1

NTA final key.

Chapter: Motion in a Plane
Topic: Projectile Motion
Idea tested: Trajectory Equation Application

Same idea in other shifts

Asked 5× in all

26 Jul 2022, Shift 1 · Q51If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is $y = 5x(1-x)$ .…MediumNumerical value
27 Jul 2022, Shift 2 · Q32A body of mass 10 kg is projected at an angle of $45^\circ$ with the horizontal. The trajectory of the body is observed to pass through a…MediumSingle correct
3 Aug 2021, Shift 2 · Q8A projectile is thrown with initial velocity of $(3\hat{i}+4\hat{j})\ \mathrm{m/s}$ . Here ' $\hat{i}$ ' is along the horizontal direction…MediumSingle correct
26 Feb 2021, Shift 2 · Q9The trajectory of a projectile in a vertical plane is $y = \alpha x - \beta x^2$ , where $\alpha$ and $\beta$ are constants and x & y are…EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.