AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Q.

The trajectory of a projectile in a vertical plane is y = ax - bx², where a and b are constants and x and y are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

a

$\frac{b^{2}}{2 a}, \tan^{- 1} (b)$

b

$\frac{a^{2}}{b} + \tan^{- 1} (2 b)$

c

$\frac{a^{2}}{4 b}, \tan^{- 1} (a)$

d

$\frac{2 a^{2}}{b}, \tan^{- 1} (a)$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

detailed_solution_thumbnail

$y = ax - {bx}^{2}; for height or y to be maximum: \frac{dy}{dx} = 0 or a - 2 bx = 0 or x = \frac{a}{2 b} y_{\max} = a (\frac{a}{2 b}) - b {(\frac{a}{2 b})}^{2} = \frac{a^{2}}{4 b} {(\frac{dy}{dx})}_{x - 0} = a = {tanθ}_{0}, where θ_{0} = angle of projection θ_{0} = \tan^{- 1} (a)$

Watch 3-min video & get full concept clarity

courses

No courses found

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

best study material, now at your finger tips!

live classes
progress tracking
24x7 mentored guidance
study plan analysis

download the app

gplay

mentor

Download the App

gplay

personalised 1:1 online tutoring

The trajectory of a projectile in a vertical plane is y = ax - bx2, where a and b are constants and x and y are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are