Q.

The equation of trajectory of projectile is given by y=x3−gx220,where x and y are in metre. The maximum range of the projectile is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

83m

b

43m

c

34m

d

38m

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

Comparing the given equation with the equation of trajectory of a projectile,y=xtanθ−gx22u2cos2θWe get, tanθ=13⇒θ=300And 2u2cos2θ=20⇒u2=202cos2θ=10cos2300=10(32)2=403Now, Rmax=u2g=403×10=43m
Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
The equation of trajectory of projectile is given by y=x3−gx220,where x and y are in metre. The maximum range of the projectile is