Home
Class 11
PHYSICS
If T is the total time of flight, h is t...

If T is the total time of flight, h is the maximum height and R is the range for horizontal motion, the x and y coordinates of projectile motion and time t are related as

A

`y = 4h (t/T)(1-t/T)`

B

`y = 4h (x/R)(1-x/R)`

C

`y = 4h (T/t)(1-T/t)`

D

`y = 4h (R/x)(1-R/x)`

Text Solution

AI Generated Solution

The correct Answer is:
To derive the relationship between the x and y coordinates of projectile motion in terms of time \( t \), total time of flight \( T \), maximum height \( h \), and range \( R \), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Basic Equations**: - The total time of flight \( T \) for a projectile is given by: \[ T = \frac{2u \sin \theta}{g} \] - The maximum height \( h \) reached by the projectile is given by: \[ h = \frac{u^2 \sin^2 \theta}{2g} \] - The range \( R \) of the projectile is given by: \[ R = u \cos \theta \cdot T \] 2. **Expressing x and y Coordinates**: - The x-coordinate at any time \( t \) can be expressed as: \[ x = u \cos \theta \cdot t \] - The y-coordinate at any time \( t \) can be expressed as: \[ y = u \sin \theta \cdot t - \frac{1}{2} g t^2 \] 3. **Substituting Time of Flight**: - From the equation for \( x \), we can express \( t \) in terms of \( x \): \[ t = \frac{x}{u \cos \theta} \] 4. **Substituting for y**: - Substitute \( t \) into the equation for \( y \): \[ y = u \sin \theta \left(\frac{x}{u \cos \theta}\right) - \frac{1}{2} g \left(\frac{x}{u \cos \theta}\right)^2 \] - Simplifying this gives: \[ y = \frac{x \tan \theta}{u \cos \theta} - \frac{g x^2}{2 u^2 \cos^2 \theta} \] 5. **Using the Maximum Height**: - We can express \( \tan \theta \) in terms of \( h \) and \( R \): \[ \tan \theta = \frac{h}{\frac{R}{2}} \quad \text{(using the relationship between height and range)} \] 6. **Final Relationship**: - After substituting and simplifying, we arrive at the final equation: \[ y = 4h \left(\frac{x}{R}\right) \left(1 - \frac{x}{R}\right) \] ### Final Result: Thus, the relationship between the x and y coordinates of projectile motion is: \[ y = 4h \left(\frac{x}{R}\right) \left(1 - \frac{x}{R}\right) \]

To derive the relationship between the x and y coordinates of projectile motion in terms of time \( t \), total time of flight \( T \), maximum height \( h \), and range \( R \), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Basic Equations**: - The total time of flight \( T \) for a projectile is given by: \[ T = \frac{2u \sin \theta}{g} ...
Promotional Banner

Topper's Solved these Questions

  • MISCELLANEOUS KINEMATICS

    CENGAGE PHYSICS ENGLISH|Exercise Linked Comprehension Type|35 Videos
  • MISCELLANEOUS KINEMATICS

    CENGAGE PHYSICS ENGLISH|Exercise Integer Type|17 Videos
  • MISCELLANEOUS KINEMATICS

    CENGAGE PHYSICS ENGLISH|Exercise Interger type|3 Videos
  • LINEAR AND ANGULAR SIMPLE HARMONIC MOTION

    CENGAGE PHYSICS ENGLISH|Exercise Single correct anwer type|14 Videos
  • MISCELLANEOUS VOLUME 2

    CENGAGE PHYSICS ENGLISH|Exercise INTEGER_TYPE|10 Videos

Similar Questions

Explore conceptually related problems

If the maximum vertical height and horizontal ranges of a projectile are same, the angle of projection will be

Under what conditions the formulae of range, time of flight and maximum height can be applied directly in case of a projectile motion ?

If the time of flight of a bullet over a horizontal range R is T, then the angle of projection with horizontal is -

Range of a projectile with time of flight 10 s and maximum height 100 m is : (g= -10 m//s ^2)

At a height of 45 m from ground velocity of a projectile is, v = (30hati + 40hatj) m//s Find initial velocity, time of flight, maximum height and horizontal range of this projectile. Here hati and hatj are the unit vectors in horizontal and vertical directions.

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

The horizontal range of projectile is 4sqrt(3) times the maximum height achieved by it, then the angle of projection is

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

If T_1 and T_2 are the times of flight for two complementary angles, then the range of projectile R is given by

The time of flight of a projectile is related to its horizontal range by the equation gT^2 = 2R . The angle of projection is

CENGAGE PHYSICS ENGLISH-MISCELLANEOUS KINEMATICS-Multiple Correct Answer Type
  1. A particle is moving along the x-axis whose position is given by x= 4...

    Text Solution

    |

  2. A particle is thrown in vertically in upward direction and passes thre...

    Text Solution

    |

  3. Ship A is located 4 km north and 3 km east of ship B. Ship A has a vel...

    Text Solution

    |

  4. An object moves with constant acceleration a. Which of the following e...

    Text Solution

    |

  5. A ball is dropped from a height of 49 m. The wind is blowing horizonta...

    Text Solution

    |

  6. An object may have

    Text Solution

    |

  7. From the top of a tower of height 200 m, a ball A is projected up with...

    Text Solution

    |

  8. A particle is projected at an angle theta =30^@ with the horizontal, w...

    Text Solution

    |

  9. A particle moves along positive branch of the curve, y = x/2, where x ...

    Text Solution

    |

  10. If T is the total time of flight, h is the maximum height and R is the...

    Text Solution

    |

  11. If acceleration is constant and initial velocity of the body is 0, the...

    Text Solution

    |

  12. The velocity - time graph of two bodies A and B is shown in figure. Ch...

    Text Solution

    |

  13. A particle starts moving along a straight line path with velocity 10m/...

    Text Solution

    |

  14. Consider a shell that has a muzzle velocity of 45ms^(-1) fired from th...

    Text Solution

    |

  15. A ball is projected from ground with speed u, at an angle theta above ...

    Text Solution

    |

  16. A boat is travelling due east at 12ms^(-1). A flag on the boat flaps a...

    Text Solution

    |

  17. A particle has initial velocity 4i + 4j ms^(-1) and an acceleration -0...

    Text Solution

    |

  18. A cubical box dimension L = 5//4 m starts moving with an acceleration ...

    Text Solution

    |

  19. A particle of mass m moves on the x- axis as follows : it starts from...

    Text Solution

    |

  20. The coordinates of a particle moving in a plane are given by x=acos pt...

    Text Solution

    |