Home
Class 12
PHYSICS
The horizontal range and miximum height ...

The horizontal range and miximum height attained by a projectile are `R and H`, respectively. If a constant horizontal acceleration `a = g//4` is imparted to the projectile due to wind, then its horizontal range and maximum height will be

Text Solution

Verified by Experts

`H^(1)=H ( :. U sin theta` remains same ) : T=T
`R^(1)=u_(A)+T+(1)/(2)aT^(2)=R+(1)/(2)(g)/(4)T^(2)=R+(1)/(2)10xx1^(2)`
`R^(1)=R+H " " H^(1)=H`
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-A (Vectors & Scalars)|25 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-A (Addition & Subtractions of Vectors)|10 Videos

  • MAGNETISM

    AAKASH SERIES|Exercise ADDITIONAL EXERCISE|22 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise QUESTION FOR DESCRIPTIVE ANSWER|7 Videos

Similar Questions

Explore conceptually related problems

If R and H are the horizontal range and maximum height attained by a projectile , than its speed of prjection is

Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.

Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

Find the angle of projection of a porjectile for which for horizontal range and maximum height are equal.

If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

What is the angle made by the trajectory of a projectile with horizontal at maximum height ?

For a projectile 'R' is range and 'H' is maximum height

Find the ratio of horizontal range and the height attained by the projectile. i.e. for theta = 45^(@)

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