Home
Class 11
PHYSICS
The trajectory of a projectile in a vert...

The trajectory of a projectile in a vertical plane is `y=sqrt(3)x-2x^(2)`. `[g=10 m//s^(2)]`

Maximum height H is :-

A

`8/3`

B

`3/8`

C

`sqrt(3)`

D

`2/sqrt(3)`

Text Solution

Verified by Experts

The correct Answer is:
B
Max. height `H=(u^(2) sin^(2) theta)/(2g),(10xx(sqrt(3)/2)^(2))/(2xx10)=3/8 m`
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS

    ALLEN|Exercise Comprehension#4|3 Videos

  • KINEMATICS

    ALLEN|Exercise Comprehension#5|6 Videos

  • KINEMATICS

    ALLEN|Exercise Comprehension#2|5 Videos

  • ERROR AND MEASUREMENT

    ALLEN|Exercise Part-2(Exercise-2)(B)|22 Videos

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN|Exercise BEGINNER S BOX-7|8 Videos

Similar Questions

Explore conceptually related problems

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Range OA is :-

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Time of flight of the projectile is :-

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Angle of projection theta is :-

The trajectory of a projectile in a vertical plane is y = ax - bx^2 , where a and b are constant 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 projectile from the horizontal are.

The trajectory of a projectile in a vertical plane is y = ax - bx^2 , where a and b are constant 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 projectile from the horizontal are.

The equation of trajectory of an oblique projectile y = sqrt(3) x - (g x^(2))/(2) . The angle of projection is

The equation of trajectory ofa projectile is y = 10x-5/9x^2 If we assume g = 10 ms^(-2) , the range of projectile (in metre) is

Trajectory of particle in a projectile motion is given as y = x - (x^(2))/(80) . Here x and y are in metre. For this projectile motion match the following with g = 10 ms^(-2) .

Trajectory of particle in a projectile motion is given as y=x- x^(2)/80 . Here, x and y are in meters. For this projectile motion, match the following with g=10 m//s^(2) . {:(,"Column-I",,"Column-II"),((A),"Angle of projection (in degrees)",(P),20),((B),"Angle of velocity with horizontal after 4s (in degrees)",(Q),80),((C),"Maximum height (in meters)",(R),45),((D),"Horizontal range (in meters)",(S),30),(,,(T),60):}

Two graphs of the same projectile motion (in the xy-plane) projected from origin are shown. X-axis is along horizontal direction and y-axis is vertically upwards. Take g = 10 ms^(-2) . , Maximum height attained from the point of projection is