Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6

Q.

Match list I with list II for a projectile.

 List - I List - II
A.For two angles θ and 90θ with same magnitude of velocity of projectione. Pi¯.Pi¯g
B.Equation of parabola of a projectile, y=PxQx2f.Maximum height = 25% of P2Q
C.Radius of curvature of path of projectile projected with a velocity (Pi¯+Qj¯)ms1 at highest pointg.Range = Maximum height
D.Angle of projection θ = Tan-1(4)h.Range is same
 ABCD
1.fhge
2.hfeg
3.egfh
4.eghf

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

2

b

1

c

3

d

4

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

For complementary angle of projection, ranges are same.

For Equation of parabola of a projectile, y=PxQx2 :

H=P24Q=25% of P2Q

Radius of curvature of path of projectile projected with a velocity (Pi¯+Qj¯)ms1 at highest point :

R=vx2g=P2g=Pi¯Pi¯g

again, θ=tan1(4HR)=tan1(4)

R=H

Watch 3-min video & get full concept clarity

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
Match list I with list II for a projectile. List - I List - IIA.For two angles θ and 90∘−θ with same magnitude of velocity of projectione. Pi¯.Pi¯gB.Equation of parabola of a projectile, y=Px−Qx2f.Maximum height = 25% of P2QC.Radius of curvature of path of projectile projected with a velocity (Pi¯+Qj¯)ms−1 at highest pointg.Range = Maximum heightD.Angle of projection θ = Tan-1(4)h.Range is same ABCD1.fhge2.hfeg3.egfh4.eghf