EASYAmity University Noida B.Tech Admissions 2025
ApplyAmong Top 30 National Universities for Engineering (NIRF 2024) | 30+ Specializations | AI Powered Learning & State-of-the-Art Facilities
Equation Of Path Of A Projectile - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
-
22 Questions around this concept.
Solve by difficulty
A projectile is given an initial velocity of , where
is along the ground and
is along the vertical. If g = 10m/s2, the equation of its trajectory is :
An initial velocity of $(2 \hat{i}+\sqrt{3} \hat{j}) \mathrm{m} / \mathrm{s}$ is given to a projectile, which $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. The equation of its trajectory is if $g=10 \mathrm{~m} / \mathrm{s}^2$.
A particle is moving according to the equation $y=\left(\sqrt{3} X-2 X^2\right)$. Then the horizontal range of project is
JEE Main 2025: Session 2 Result Out; Direct Link
JEE Main 2025: College Predictor | Marks vs Percentile vs Rank
New: JEE Seat Matrix- IITs, NITs, IIITs and GFTI | NITs Cutoff
Concepts Covered - 1
Equation of path of a projectile
$
y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}
$
or-
$
y=x \tan \theta\left(1-\frac{x}{R}\right)
$
Where, R is the horizontal range of the projectile.
It is equation of parabola, So the trajectory path of the projectile is parabolic in nature
$g \rightarrow \quad$ Acceleration due to gravity
$u \rightarrow$ initial velocity
$\theta=$ Angle of projection
Study it with Videos
Equation of path of a projectile"Stay in the loop. Receive exam news, study resources, and expert advice!"
Get Answer to all your questions
JEE Main Articles
Back to top