Home
Class 11
PHYSICS
A body is thrown at an angle theta0 with...

A body is thrown at an angle `theta_0` with the horizontal such that it attains a speed equal to `sqrt((2)/(3))` times the speed of projection when the body is at half of its maximum height. Find the angle `theta_0`.

Text Solution

AI Generated Solution

To solve the problem, we need to find the angle of projection \( \theta_0 \) when a body is thrown at this angle and reaches a speed equal to \( \sqrt{\frac{2}{3}} \) times the speed of projection at half of its maximum height. ### Step-by-Step Solution: 1. **Identify the Variables:** - Let \( u \) be the initial speed of projection. - The angle of projection is \( \theta_0 \). - The maximum height \( H \) is given by the formula: ...
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS-2

    CENGAGE PHYSICS ENGLISH|Exercise Solved Examples|7 Videos

  • KINEMATICS-2

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 5.1|15 Videos

  • KINEMATICS-1

    CENGAGE PHYSICS ENGLISH|Exercise Integer|9 Videos

  • KINETIC THEORY OF GASES

    CENGAGE PHYSICS ENGLISH|Exercise Compression|2 Videos

Similar Questions

Explore conceptually related problems

If a body is projected with an angle theta to the horizontal, then

A particle is projected at an angle theta with an initial speed u .

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

A projectile is thrown with velocity v at an angle theta with the horizontal. When the projectile is at a height equal to half of the maximum height,. The velocity of the projectile when it is at a height equal to half of the maximum height is.

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is .

The horizontal range of a projectile is 4 sqrt(3) times its maximum height. Its angle of projection will be

The horizontal range of a projectile is 4 sqrt(3) times its maximum height. Its angle of projection will be

A stone is thrown at an angle theta to the horizontal reaches a maximum height H. Then the time of flight of stone will be:

A stone is thrown at an angle theta to the horizontal reaches a maximum height H. Then the time of flight of stone will be:

A particle is projected with a speed u at an angle theta to the horizontal. Find the radius of curvature. At the point where the particle is at a highest half of the maximum height H attained by it.