Home
Class 12
PHYSICS
The equation of trajectory of a projecti...

The equation of trajectory of a projectile thrown from a level ground near the surface of earth is given by y = ax `- bx^(2)` , with y axis in vertical direction and x-axis in horizontal direction .a and b are constants. Then

A

The range of the projectile is ` a/b`

B

At ` x = a/(2b)` , the velocity of projectile becomes zero

C

The maximum height attained by projectile is ` a^(2)/(4b)`

D

The angle of projection is ` tan^(-1)(a)`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH INSTITUTE|Exercise Assignement section -D (Linked Comprehension)|8 Videos

  • MOTION IN A PLANE

    AAKASH INSTITUTE|Exercise Assignement section -E (Assertion-Reason)|3 Videos

  • MOTION IN A PLANE

    AAKASH INSTITUTE|Exercise Assignement section -B Objective (one option is correct)|24 Videos

  • MOCK_TEST_17

    AAKASH INSTITUTE|Exercise Example|15 Videos

  • MOTION IN A STRAIGHT LINE

    AAKASH INSTITUTE|Exercise ASSIGNMENT (SECTION - D)|15 Videos

Similar Questions

Explore conceptually related problems

The equation of trajectory of a projectile thrown from a point on the ground is y=(x-x^(2)/40) m. If g=10ms^(-2) The maximum height reached is

The equation of a trajectory of a projectile is y=8x-4x^(2) . The maximum height of the projectile is ?

Knowledge Check

  • The equation of a projectile is y = ax - bx^(2) . Its horizontal range is

    A
    `(a)/(b)`
    B
    `(b)/(a)`
    C
    a + b
    D
    b - a

  • 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.

    A
    `(b^2)/(2 a), tan^-1 (b)`
    B
    `(a^2)/(b), tan^-1 (2b)`
    C
    `(a^2)/(4 b), tan^-1 (a)`
    D
    `(2 a^2)/(b), tan^-1 (a)`

    • Similar Questions

    Explore conceptually related problems

    The equation of path of a particle projected obliquely from the ground under gravity is given by y=2x-x^(2) where 'x' and ' y' are in meter.The "x" and "y" axes are in horizontal and vertical directions respectively.The speed of projection is

    Find the equation of trajectory, time of flight , maximum height and horizontal range of a projectile when projected at an angle theta with the vertical direction .

    Trajectory of particle in projectile motion is y=(x-x^(2)//80) , x and y are in metre. Projectile range is on horizontal plane.

    The equation of a straight line which cuts off an intercept of 5 units on negative direction of y-axis and makes an angle of 120^(@) with the positive direction ofx-axis is

    [" The equation of trajectory of a projectile is "],[y=ax-bx^(2)" where "x" and "y" axes are along "],[" horizontal and vertical directions "],[" respectively.The range of the projectile is "]

    Find the equation of the line cutting off an intercept of length 2 from the negative direction of the axis of y and making an angle of 120^@ with the positive direction of X-axis.

    Home
    Profile