Home
Class 11
PHYSICS
Find the angle of projection of a projec...

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

A

`theta=45^(@)`

B

`theta=tan^(-1)(1/4)`

C

`theta=tan^(-1)(4)`

D

`theta=tan^(-1)(2)`

Text Solution

Verified by Experts

The correct Answer is:
c
Given `H=R, (u^(2) sin^(2) theta)/(2g)=(2u^(2) sin theta cos theta)/g`
`implies tan theta=4 implies theta=tan^(-1)(4)`
Promotional Banner

Topper's Solved these Questions

  • MOTION IN TWO DIMENSION

    A2Z|Exercise AIIMS Questions|15 Videos

  • MOTION IN TWO DIMENSION

    A2Z|Exercise Chapter Test|29 Videos

  • MOTION IN TWO DIMENSION

    A2Z|Exercise Assertion Reasoning|11 Videos

  • MOCK TEST

    A2Z|Exercise Motion With Constant Acceleration|15 Videos

  • NEWTONS LAWS OF MOTION

    A2Z|Exercise Chapter Test|30 Videos

Similar Questions

Explore conceptually related problems

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

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

The angle of projection at which the horizontal range and maximum height of projectile are equal is

Find the angle of projection at which horizontal range and maximum height are equal.

Find the angle of projection for a projectile motion whose rang R is (n) time the maximum height H .

A body is thrown at an angle alpha to the horizontal with initial velocity v_0 .Assuming air drag to be negligible , find (i)the time of motion, (ii)maximum height of ascent and the horizontal range, (iii)the angle of projection at which they are equal , (iv)the angle of projection at which the horizontal range is maximum and the angle at which the vertical range is maximum.

For given value of u, there are two angles of projection for which the horizontal range is the same. the sum of the maximum heights for these two angles is

Assertion: For projection angle tan^(-1)(4) , the horizontal range and the maximum height of a projectile are equal. Reason: The maximum range of projectile is directely proportional to square of velocity and inversely proportional to acceleration due to gravity.

If the angle of projection theta corresponds to horizontal range being equal to the maximum height then tantheta equals :

A2Z-MOTION IN TWO DIMENSION-NEET Questions
  1. Two particles of mass M and m are moving in a circle of radii R and r....

    Text Solution

    |

  2. A particle (A) is dropped from a height and another particles (B) is t...

    Text Solution

    |

  3. From a 10 m high building a stone 'A' is dropped, and simultaneously a...

    Text Solution

    |

  4. Two boys are standing at the ends A and B of a ground, where AB=a. The...

    Text Solution

    |

  5. A stone tied to the end of string 1m long is whirled in a horizontal c...

    Text Solution

    |

  6. A car runs at a constant speed on a circular track of radius 100m, tak...

    Text Solution

    |

  7. For angles of projection of a projectile at angle (45^(@) - theta) and...

    Text Solution

    |

  8. A paricle starting from the origin (0,0) moves in a straight line in (...

    Text Solution

    |

  9. A particle moves in a circle of radius 5 cm with constant speed and ti...

    Text Solution

    |

  10. A missile is fired for maximum range with an initial velocity of 20m//...

    Text Solution

    |

  11. A projectile is fired at an angle of 45^(@) with the horizontal. Eleva...

    Text Solution

    |

  12. Find the angle of projection of a projectile for which the horizontal ...

    Text Solution

    |

  13. A particle has initial velocity (2hati+3hatj) and acceleration (0.3hat...

    Text Solution

    |

  14. The velocity of a projectile at the initial point A is (2hati+3hatj) m...

    Text Solution

    |

  15. A ship A is moving Westwards with a speed of 10kmh^(-1) and a ship B 1...

    Text Solution

    |

  16. The x and y coordinates of the particle at any time are x=5t-2t^(2) an...

    Text Solution

    |