PROJECTILE MOTION Projectile motion is a motion in
- Slides: 30
PROJECTILE MOTION
Projectile motion is a motion in which an object is thrown near the earth’s surface, and it moves along the curved path under the action of gravity only.
Motion of a snow boarder descending from a slope
Motion of a rider performing a bike stunt
Motion of water jets coming out from nozzles
Assumptions used in projectile motion 1. The effect due to air resistance is negligible. 2. The effect due to curvature of the earth is negligible. 3. The effect due to rotation of the earth is negligible. 4. The acceleration due to gravity is constant over the range of motion.
Idealized model of projectile motion v H θ R
Projectile Any object which is projected in the air is called as projectile. Projectile v H θ R
Point of projection The point from which the object is projected in air is called as point of projection. Point of projection v H θ R
Velocity of projection The velocity with which an object is projected in air is called as velocity of projection. Velocity of projection v H θ R
Angle of projection The angle with the horizontal at which an object is projected in air is called as angle of projection. v Angle of projection H θ R
Trajectory The parabolic path followed by a projectile in air is called its trajectory. v Trajectory H θ R
Time of flight Time taken by the projectile to cover the entire trajectory is called as time of flight. v H T θ Time of flight R
Maximum height of projectile It is the maximum vertical distance travelled by the projectile from the ground level during its motion. Maximum height v H θ R
Horizontal range of projectile It is the horizontal distance travelled by the projectile during entire motion. v H θ R Horizontal range
Is horizontal and vertical motions are interdependent? • Both pink and yellow balls are falling at the same rate. • Yellow ball is moving horizontally while it is falling have no effect on its vertical motion. • It means horizontal and vertical motions are independent of each other.
Analysis of projectile motion Motion diagram of a kicked football
Analysis of projectile motion Projectile motion + Horizontal motion �� �� v i�� Vertical motion vi v i�� �� o v i�� Frame - 1 v i�� o ��
Analysis of projectile motion Projectile motion Vertical motion + Horizontal motion �� �� v 1 v 1�� �� o Frame - 2 v 1�� o ��
Analysis of projectile motion Projectile motion Vertical motion + Horizontal motion �� �� v 2�� �� o Frame - 3 v 2�� o ��
Analysis of projectile motion Projectile motion Vertical motion + Horizontal motion �� �� v 3�� �� o Frame - 4 v 3�� o ��
Analysis of projectile motion Projectile motion Vertical motion + Horizontal motion �� �� v f�� �� o o Frame - 5 v f�� vf v f�� ��
Important conclusion Projectile motion is a combination of two motions: a) Horizontal motion with constant velocity b) Vertical motion with constant acceleration
Equation of trajectory The horizontal distance travelled by the projectile in time ‘t’ along �� -axis is, x 1 x y = v isin θ × − g vicos θ 2 x = v i�� t x = (v icos θ)t x t= vi cos θ y = tan θ x − … …. . (1) The vertical distance travelled by the projectile in time ‘t’ along �� -axis is, 1 y = v i �t�− g t 2 2 1 2 … …. . (2) y = (vi sin θ)t − g t 2 Substituting the value of ‘t’ from eq. (1) in eq. (2) we get, g 2 x 2 vi 2 cos 2 θ This is the equation of trajectory of a projectile. As vi , θ and g are constant for a given projectile, we can write g tan θ = α & =β 2 vi 2 cos 2 θ y = αx − βx 2 Thus y is a quadratic function of x. Hence the trajectory of a projectile is a parabola.
Equation for time of flight The vertical distance travelled by the projectile in time ‘t’ along �� -axis is, 1 y = v i y t − g t 2 2 For entire motion, t = T and y = 0 1 0 = (vi sin θ)T − g T 2 2 1 g T 2 = (vi sin θ)T 2 2 vi sin θ T= g This is the equation for time of flight. For symmetrical parabolic path, time of ascent is equals to time of descent. TA = TD = T 2 vi sin θ g
Equation for horizontal range The equation of trajectory of projectile is given by, y = tan θ x − g 2 x 2 vi 2 cos 2 θ For entire motion, y = 0 and x = R 0 = tan θ R − g 2 R 2 vi 2 cos 2 θ g 2 = tan θ R R 2 vi 2 cos 2 θ sin θ g ×R= 2 vi 2 cos 2 θ cos θ vi 2 (2 sin θ cos θ) R= g vi 2 sin 2θ R= g This is the equation for horizontal range. For a given velocity of projection the range will be maximum when sin 2θ = 1, 2θ = 90° & θ = 45° Thus range of the projectile is maximum if it is projected in a direction inclined to the horizontal at an angle of 45°.
Two angles of projection for the same horizontal range The equation for horizontal range is given by, vi 2 sin 2θ R= g … …. . (1) Replacing θ by 90° − θ, From eq. (1) & (2) we say that, R = R′ Thus horizontal range of projectile is same for any two complementary angles i. e. θ and 90° − θ. �� vi 2 sin 2(90° − θ) R′ = g 30° vi 2 sin(180° − 2θ) R′ = g vi 2 sin 2θ R′ = g 60° … …. . (2) o R = R′ ��
Equation for maximum height The third kinematical equation for vertical motion is, Case-1: Angle of projection is 45°. 2 vf� = vi 2 y − 2 g y � At maximum height: vf y = 0 & �� =H 0 = vi sin θ 2− 2 g. H 2 g H = vi sin θ H= 2 g 2 2 This is the equation for maximum height. 2 vi sin 45° H= 2 g v 1 2 H= 2 g 2 i 2 ⇒ Hmax = vi 2 4 g Case-2: Angle of projection is 90°. vi sin 90° H= 2 g vi 2 1 H= 2 g 2 2 ⇒ H= vi 2 2 g
Thank you
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