(C) The horizontal position is given by $x = u_x t$. Given $x = 40\,m$ at $t = 2\,s$,we have $40 = u_x \times 2$,so $u_x = 20\,m/s$.The vertical posit

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The horizontal position is given by $x = u_x t$. Given $x = 40\,m$ at $t = 2\,s$,we have $40 = u_x \times 2$,so $u_x = 20\,m/s$.The vertical position is given by $y = u_y t - \frac{1}{2} g t^2$. Given $y = 50\,m$ at $t = 2\,s$ and $g = 10\,m/s^2$,we have $50 = u_y(2) - \frac{1}{2}(10)(2)^2$.$50 = 2u_y - 20$,which implies $2u_y = 70$,so $u_y = 35\,m/s$.The angle of projection $\theta$ is given by $\tan \theta = \frac{u_y}{u_x}$.$\tan \theta = \frac{35}{20} = \frac{7}{4}$.Therefore,$\theta = \tan^{-1}(\frac{7}{4})$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

Difficult

The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)

JEE MAIN 2014 All JEE MAIN

A
$\tan^{-1}(\frac{2}{3})$
B
$\tan^{-1}(\frac{3}{2})$
C
$\tan^{-1}(\frac{7}{4})$
D
$\tan^{-1}(\frac{4}{5})$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

3-2.Motion in Plane

Subtopic

Horizontal Projectile Motion

Previous Year

JEE MAIN 2014 Paper All JEE MAIN years →

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are $5 \text{ s}$ and $10 \text{ s}$, respectively, then the value of $R$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

DifficultJEE MAIN 2026

View Solution

If the range of a projectile is twice its maximum height,what is the range?

Difficult

View Solution

$A$ body is projected horizontally from the top of a tall tower with a velocity of $30 \,ms^{-1}$. At time $t_1$, its horizontal and vertical components of the velocity are equal and at time $t_2$, its horizontal and vertical displacements are equal. Then $t_2-t_1$ is (take, $g=10 \,ms^{-2}$ ) (in $\,s$)

MediumAP EAMCET 2018

View Solution

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Vertical component of velocity is zero $(a)$ Tangent to the parabolic path

$(2)$ Linear velocity $(b)$ Maximum height point of the projectile trajectory

Column-$I$	Column-$II$
$(1)$ Vertical component of velocity is zero	$(a)$ Tangent to the parabolic path
$(2)$ Linear velocity	$(b)$ Maximum height point of the projectile trajectory

Easy

View Solution

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$, where $h$ is the height, $x$ is the horizontal distance, and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

DifficultAP EAMCET 2001

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)

Similar Questions

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are $5 \text{ s}$ and $10 \text{ s}$, respectively, then the value of $R$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

If the range of a projectile is twice its maximum height,what is the range?

Match Column-$I$ with Column-$II$. Column-$I$ Column-$II$ $(1)$ Vertical component of velocity is zero $(a)$ Tangent to the parabolic path $(2)$ Linear velocity $(b)$ Maximum height point of the projectile trajectory

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$, where $h$ is the height, $x$ is the horizontal distance, and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Vertical component of velocity is zero $(a)$ Tangent to the parabolic path

$(2)$ Linear velocity $(b)$ Maximum height point of the projectile trajectory