(C) The horizontal velocity $u_x = u \cos \theta$ is equal to the velocity acquired from rest with acceleration $a = 3 \, ms^{-2}$ for time $t = 0.5 \

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

(C) The horizontal velocity $u_x = u \cos \theta$ is equal to the velocity acquired from rest with acceleration $a = 3 \, ms^{-2}$ for time $t = 0.5 \, minutes = 30 \, s$.$u_x = a \times t = 3 \times 30 = 90 \, ms^{-1}$.The maximum height $H$ is given by $H = \frac{u_y^2}{2g}$,where $u_y = u \sin \theta$.$80 = \frac{(u \sin \theta)^2}{2 \times 10} \implies (u \sin \theta)^2 = 1600 \implies u_y = 40 \, ms^{-1}$.The angle of projection $\theta$ is given by $\tan \theta = \frac{u_y}{u_x}$.$\tan \theta = \frac{40}{90} = \frac{4}{9}$.Therefore,$\theta = \tan^{-1}(4/9)$.

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

Difficult

$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)

A
$\tan^{-1}(3)$
B
$\tan^{-1}(3/2)$
C
$\tan^{-1}(4/9)$
D
$\sin^{-1}(4/9)$

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

$A$ bullet is fired with a velocity $u$ making an angle of $60^{\circ}$ with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is

MediumWBJEE 2009

View Solution

$A$ ball is thrown at $30^{\circ}$ with the horizontal from the top of a roof $20 \,m$ high with a speed of $13 \,ms^{-1}$. At what distance from the throwing point will the ball, once again, be at a height of $20 \,m$ from the ground (in $\,m$)? $(g = 10 \,ms^{-2})$

MediumAP EAMCET 2021

View Solution

Two projectiles are projected at $30^{\circ}$ and $60^{\circ}$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

MediumAIIMS 2001

View Solution

$A$ stone is projected from the ground at $t = 0$. At the time of projection,the horizontal and vertical components of velocity are $10\, m/s$ and $20\, m/s$ respectively. Find the time at which the magnitudes of tangential and normal acceleration will be equal $(g = 10\, m/s^2)$ [neglect air friction].

Difficult

View Solution

$A$ particle is projected at an angle of $60^{\circ}$ with the horizontal from the ground with a velocity $10 \sqrt{3} \ m/s$. The angle between the velocity vector after $2 \ s$ and the initial velocity vector is $(g = 10 \ m/s^2)$. (in $^{\circ}$)

EasyAP EAMCET 2018

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

$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)

Similar Questions

$A$ bullet is fired with a velocity $u$ making an angle of $60^{\circ}$ with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is

$A$ ball is thrown at $30^{\circ}$ with the horizontal from the top of a roof $20 \,m$ high with a speed of $13 \,ms^{-1}$. At what distance from the throwing point will the ball, once again, be at a height of $20 \,m$ from the ground (in $\,m$)? $(g = 10 \,ms^{-2})$

Two projectiles are projected at $30^{\circ}$ and $60^{\circ}$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

$A$ particle is projected at an angle of $60^{\circ}$ with the horizontal from the ground with a velocity $10 \sqrt{3} \ m/s$. The angle between the velocity vector after $2 \ s$ and the initial velocity vector is $(g = 10 \ m/s^2)$. (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module