1. PHYSICS PROJECT PRESENTED BY : Shahrukh Soheil Rahman,11 A1 PROJECTILE MOTION General Information, Definition Of Terms, Derivation Of Equations and Examples
2. The undersigned wishes to thank Mr. Debangshu Deb,PGT,Physics for his guidance and suggestions,without which this project wouldn’t have been successful.The undersigned also wishes to thank her parents for their valuable time and support and last but not least her friends for their valuable feedback. Shreya Choudhury,11 A1 ACKNOWLEDGEMENTS
3. PROJECTILE : An object thrown with initial velocity and which then allowed to move under the action of gravity alone is called a projectile.
4. Before we begin learning more about the projectiles,lets know the definitions of some terms associated with projectile motion. TRAJECTORY : The path followed by a projectile during its flight is known as trajectory TIME OF FLIGHT : The time taken by the projectile to complete its trajectory is called time of flight.Denoted by- T HORIZONTAL RANGE : The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground is called horizontal range.Denoted by- R MAXIMUM HEIGHT : The maximum vertical distance covered by the projectile during its journey is called the maximum height attained by the projectile.Denoted by- H IMPORTANT TERMS
5. PROJECTILES An object thrown with initial velocity and then allowed to move under the action of gravity alone is known as projectile Few examples of projectiles are: 1: A bomb released from an airplane . 2: A bullet shot from a gun . 3: A javelin thrown . Etc The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along the horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity. It was Galileo who first stated this independency of the horizontal and the vertical components of the projectile motion in his Dialogue On The Great World Systems(1632)
6. Consider a projectile thrown horizontally with an initial velocity u from the top of a tower of height h as shown in the figure. As it moves, it covers the horizontal distance due to the uniform horizontal velocity, u and vertically downward distance because of constant acceleration due to gravity, g . Let the horizontal distance travelled in time t = x(t) = t And the vertical distance travelled by it = y(t) = t The motion is in plane and the equations describing this motion are given by, … (1) … (2) EQUATION OF PATH OF A PROJECTILE The projectile moves along the path OPA. Time taken to reach A = t
7. For horizontal motion of a projectile [ Since there is no change in velocity in the horizontal direction ] Therefore from equation (1) we get, or … (3) For vertical motion of a projectile [ Acceleration due to gravity ] Therefore from equation (2) we get, … (4) EQUATION OF PATH OF A PROJECTILE
8. Substituting the value of t from equation (3) in equation (4),we get, or Where is a constant EQUATION OF PATH OF A PROJECTILE From the above we get that the path of a projectile is parabolic
10. Let a projectile be thrown with initial velocity u at an angle 0 with the horizontal direction. Thus, horizontal component of the initial velocity, And vertical component of the initial velocity, The projectile travels in the horizontal direction with a constant velocity, The projectile travels in the vertical direction with a constant velocity of, The vertical component of the velocity gradually reduces to zero at the highest point and then the projectile moves downward to fall on to the ground. The trajectory of the projectile is represented by OAB . Projectile Fired At An Angle With The Horizontal
11. since and [ since projectile is not accelerated horizontally as no force acts on it in horizontal direction ] Therefore we get, or Distance travelled by the projectile in the vertical direction in time t is given by ...(1) Projectile Fired At An Angle With The Horizontal Let at any instant of time t , the projectile be at point P . Let x and y be the horizontal and vertical distances travelled by the projectile in time t . The distance travelled by the projectile in the horizontal direction in time t is given by,
12. since and [ since g acts vertically downwards ] Therefore we get ...(2) Putting the value of t from equation (1) and equation (2) we get, Projectile Fired At An Angle With The Horizontal
13. For vertical upward motion which is along the Y- axis we have, Here At the highest point, Therefore equation (1) becomes … .(1) Maximum Height, H The maximum vertical distance travelled by a projectile during its journey is called the maximum height attained by the projectile. It is denoted by H.
14. We know that That is the vertical distance travelled by the projectile in time t Let t = T = Time Of Flight [ Net displacement in y-direction is zero] Here y = 0 … ..(1) From equation (1) we get Time Of Flight, T The total time taken by the projectile from the point of projection till it hits the horizontal plane having point of projection is called time of flight. It is denoted by T.
15. For horizontal motion we have … .(1) Here we have X = R, range of projectile t = T, time of flight Therefore equation (1) becomes But we know that Solving it we finally get Horizontal Range, R The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits is called horizontal range. It is denoted by R
17. The range of the projectile is given by or Thus the range of the projectile will be same for two angles of projections with the horizontal. The angles are, and TWO ANGLES OF PROJECTION FOR THE SAME RANGE For example if a projectile is projected at an angle of 30 degrees with the horizontal, then the range of the projectile will also be same if it is projected at an angle of (90-30)=60 degrees with the horizontal.
18. PROJECTILE THROWN VERTICALLY UPWARDS We know For We know If a projectile is thrown vertically upwards then the angle with the horizontal is 90 degrees. Hence the various quantities are reduced as follows,
19. PROJECTILE FIRED AT AN ANGLE WITH THE VERTICAL Let a projectile be fired with an initial velocity u at an angle with the vertical The components of the velocity vector are (i) along OY (II) along OX Projection of a projectile at an angle theta with the vertical is equivalent to the projectile projected at an angle ninety minus theta with the horizontal. So by substituting the value in the various expressions we have derived so far, we get the corresponding expressions for the projectile projected.
20. PROJECTILE FIRED AT AN ANGLE WITH THE VERTICAL MAXIMUM HEIGHT TIME OF FLIGHT
21. PROJECTILE FIRED AT AN ANGLE WITH THE VERTICAL HORIZONTAL RANGE VELOCITY AT ANY TIME t
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23. Shreya Choudhury SOFTWARES USED Microsoft PowerPoint 2007 Microsoft Paint Snipping Tool CREDITS SLIDE SHOW DESIGNED BY