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

A particle is projected from level ground. Assuming projection point as origin, x-axis along horizontal and y-axis along vertically upwards. If particle moves in x-y plane and its path is given by $y = a x - b x^{2}$ where $a, b$ are positive constants. Then match the physical quantities given in column-I with the values given in column – II (g in column-II is acceleration due to gravity).
COLUMN-I COLUMN-II
I Horizontal component of velocity P $\frac{a}{b}$
II Time of flight Q $\frac{a^{2}}{4 b}$
III Maximum height R $\sqrt{\frac{g}{2 b}}$
IV Horizontal range S $\sqrt{\frac{2 a^{2}}{b g}}$
T $\frac{2 a}{b}$
Now match the given columns and select the correct option forth the codes given below.
Codes:

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

$I - S, I I - P, I I I - R, I V - Q$

$I - R, I I - S, I I I - Q, I V - P$

$I - P, I I - Q, I I I - R, I V - s$

$I - Q, I I - R, I I I - Q, I V - P$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Equation of path is given as $y = a x - b x^{2}$
Comparing it with standard equation of projectile,
$y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$
$\tan θ = a, \frac{g}{2 u^{2} \cos^{2} θ} = 6$
Horizontal component of velocity = $u \cos θ = \sqrt{\frac{g}{2 b}}$
Time of flight , $T = \frac{2 u \sin θ}{g} = \frac{2 (u \cos θ) \tan θ}{g}$
$T = \frac{2 (\sqrt{\frac{g}{2 b}}) a}{g}$ = $\sqrt{\frac{2 a^{2}}{b g}}$
Maximum height, $H = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{{(u \cos θ \tan θ)}^{2}}{2 g}$
$∴ H = \frac{a^{2}}{4 b}$
Horizontal range, $R = \frac{u^{2} \sin 2 θ}{g} =$ $\frac{2 [\sqrt{\frac{g}{2 b}} a] [\sqrt{\frac{g}{2 b}}]}{g}$
$∴ R = \frac{a}{b}$

Watch 3-min video & get full concept clarity

courses

No courses found

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!