LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2024
JEE MAIN 2025 Foundation Course
NEET 2024 Crash Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
25 views
in General by (48.1k points)
closed by
The shape of the projectile is
1. a straight line
2. curvilinear motion
3. a parabola
4. a hyperbola

1 Answer

0 votes
by (35.4k points)
selected by
 
Best answer
Correct Answer - Option 3 : a parabola

Concept:

Equation of projectile will be of the form

y = Ax + Bx2

This is the equation of the parabola

Derivation:

Let the projectile velocity be U with an angle \(\begin{array}{l} \theta \\ \end{array}\) from the horizontal

Horizontal component of velocity

\(\begin{array}{l} \mathop u\nolimits_x = u\cos \theta \\ \end{array}\)

vertical component of velocity

\(\begin{array}{l} \mathop u\nolimits_y = u\sin \theta - gt\\ \end{array}\)

\(x = u\cos \theta t\)    (1)

\(y = u\sin \theta t - \frac{1}{2}g\mathop t\nolimits^2 \)     (2)

substituting t from (1) in (2), we get

\(\begin{array}{l} y = x\tan \theta - \frac{{g\mathop x\nolimits^2 }}{{2\mathop u\nolimits^2 \mathop {\cos }\nolimits^2 \theta }}\\ \end{array}\)

This is the equation of the parabola

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...