Home
Class 12
PHYSICS
If R is the horizontal range for theta ...

If R is the horizontal range for `theta` inclination and h is the maximum height reached by the projectile, show that the maximum ragne is given by `(R^(2))/(8h) + 2h`.

Text Solution

Verified by Experts

We know that horizontal range .
`R = (u^(@) sin 2theta)/(g)` and maximum height , `h = (u^(2) sin^(2) theta)/(2g)`
`therefore (R^(2))/(8h) +2h = ([(u^(2)sin 2 theta)/(g)]^(2))/(8[(u^(2) sin^(2) theta)/(2g)]) + 2 [(u^(2) sin^(2)theta)/(2g)]`
` = (u^(4) (2 sin theta cos theta)^(2))/(g^(2) xx 8 (u^(2) sin theta)/(2g)) + (u^(2) sin^(2) theta)/(g)`
` = (u^(2))/(g) (cos^(2) theta + sin^(2) theta) = (u^(2))/(g)= R_(max)`
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise SHORT ANSWER TYPE QUESTIONS|6 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise VERY SHORT ANSWER TYPE QUESTIONS|10 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-3 (Circular Motion)|7 Videos

  • MOTION IN A STRAIGHT LINE

    AAKASH SERIES|Exercise very Short answer type question|15 Videos

Similar Questions

Explore conceptually related problems

If R is the maximum horizontal range of a particle, then the greatest height attained by it is :

Find the ratio of maximum horizontal range and the maximum height attained by the projectile. i.e. for theta_(0) = 45^(@)

The horizontal range of a projectile is 2 sqrt(3) times its maximum height. Find the angle of projection.

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

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

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

Two particles projected form the same point with same speed u at angles of projection alpha and beta strike the horizontal ground at the same point. If h_1 and h_2 are the maximum heights attained by the projectile, R is the range for both and t_1 and t_2 are their times of flights, respectively, then

Two particles projected form the same point with same speed u at angles of projection alpha and beta strike the horizontal ground at the same point. If h_1 and h_2 are the maximum heights attained by the projectile, R is the range for both and t_1 and t_2 are their times of flights, respectively, then incorrect option is :

Show that the maximum height reached by a projectile launched at an angle of 45^@ is one quarter of its range.

The horizontal range (R ) of a projectile becomes (R + 2 H) from R due to a wind in horizontal direction. Here H is the maximum height reached by the projectile. What constant horizontal acceleration is imparted by the wind ?