1
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The position vector of a moving body at any instant of time is given as $\overrightarrow{\mathrm{r}}=\left(5 \mathrm{t}^2 \hat{i}-5 \mathrm{t} \hat{j}\right) \mathrm{m}$. The magnitude and direction of velocity at $t=2 s$ is,

A
$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with - ve Y axis
B
$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve $X$ axis
C
$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve $X$ axis
D
$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with $-$ ve $Y$ axis
2
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An object of mass ' m ' is projected from origin in a vertical xy plane at an angle $45^{\circ}$ with the $\mathrm{x}-$ axis with an initial velocity $\mathrm{v}_0$. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]

A
$\frac{m v_o{ }^3}{2 \sqrt{2} g}$ along negative $z$-axis
B
$\frac{m v_o^3}{2 \sqrt{2} g}$ along positive $z$-axis
C
$\frac{m v_o^3}{4 \sqrt{2} g}$ along positive $z$-axis
D
$\frac{m v_o^3}{4 \sqrt{2} g}$ along negative z-axis
3
JEE Main 2025 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A ball of mass 100 g is projected with velocity $20 \mathrm{~m} / \mathrm{s}$ at $60^{\circ}$ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is

A
20 J
B
5 J
C
15 J
D
zero
4
JEE Main 2024 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The angle of projection for a projectile to have same horizontal range and maximum height is :

A
$$\tan ^{-1}\left(\frac{1}{2}\right)$$
B
$$\tan ^{-1}(2)$$
C
$$\tan ^{-1}\left(\frac{1}{4}\right)$$
D
$$\tan ^{-1}(4)$$
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