Free Fall and Projectile motion

Transcription

1 Free Fall and Projectile motion 1. Introduction Free fall is the motion of a fallin body under the effect of the ravitational field only. Projectile motion is the motion of a body thrown with an initial velocity and continues its motion under the effect of the ravitational field only. So, as you can see, free fall can be considered as a special case of the projectile motion. 2. Motion with constant acceleration 2.1. Velocity vector Since a is a constant vector: a = d v dt d v = a dt d v = a dt (1 but a = const, then interatin equation 1, we et: v = a t + c (2 where c is a constant vector. At t = 0: v = v 0, where v 0 is the initial velocity vector (i.e. at t = 0. Substitutin this condition in equation 2, we et: v 0 = a 0 + c c = v 0 (3 So: v = a t + v 0 ( Position vector v = d r dt d r = v dt 1 d r = v dt

2 Substitutin equation 4: d r = ( a t + v 0 dt r = 1 2 a t2 + v 0 t + r 0 (5 where r 0 is the initial position vector (i.e. at t = 0, and is obtained by the same manner we derived the initial velocity vector v 0. So, for a motion with a constant acceleration: a = const v = a t + v 0 (6 r = 2 1 a t2 + v 0 t + r 0 In Cartesian coordinates, the position vector: x = 1 2 a x t 2 + v 0x t + x 0 r y = 2 1 a y t 2 + v 0y t + y 0 (7 z = 1 2 a z t 2 + v 0z t + z 0 3. Free fall of bodies 3.1. Acceleration The acceleration of the fallin body is the acceleration due to ravity, which is of manitude and of direction vertical downward. Hence we can write: a = k ( Velocity a = d v dt (9 d v = a dt d v = a dt = ( k dt v = t k + v0 (10 Since the motion is vertically downward, that is on the z-axis, then v 0 = v 0z k. So equation 10 becomes: v = ( t + v 0z k (11 2

3 3.3. Position vector v = d r ] dt d r = v dt = [( t + v 0z k dt ( r = 1 2 t2 + v 0z t k + r0 (12 Aain, since the motion is on the z-axis, then r 0 = z 0 k. So, equation 12 becomes: r = ( 2 1 t2 + v 0z t + z 0 k (13 Since the motion is alon the z-axis, we can omit the unit vector k from the above expressions of a, v, and r, and consider z instead of r. Hence the free fall equations are: a = v = t + v 0 (14 z = 2 1 t2 + v 0 t + z 0 Fiure 1. Note that v 0 (which stands for v 0z, for the initial velocity has only the z-component, and z 0, are alebraic quantities, which means that they can be positive or neative. We can define the z-axis to be vertical downwards, instead of vertical upwards, then the expression of the acceleration vector chanes to: a = k (15 Then the above equations of motion can be rewritten on this axis as: a = v = t + v 0 (16 z = 1 2 t2 + v 0 t + z 0 3

4 Aain, v 0 and z 0, are alebraic quantities, which means that they can be positive or neative. 4. Projectile motion In the projectile motion, the body is launched with an initial velocity v 0 at an anle θ with the horizontal and then continued to move under the effect of ravity only. So it moves with a constant acceleration, which is the acceleration due to the ravity. Fiure 2. The motion of the projectile lies in a vertical plane. We can take this plane to be xy-plane (the x-axis is the horizontal axis, and the y-axis the vertical axis. The trajectory of a projectile is, in eneral, parabolic, as we shall see later in the followin derivation. Since this is a motion with a constant acceleration, we can use the equations of motion which are derived previously. We will rewrite them: a = const = j v = a t + v 0 = t j + v0 (17 r = 2 1 a t2 + v 0 t + r 0 = 1 2 t2 j + v0 t + r 0 At t = 0, the projectile starts from the oriin, therefore: Initial velocity components: (Fiure 2 v 0x = v 0 cos θ i v 0 v 0y = v 0 sin θ j v 0z = 0 The position vector: r 0 = 0 (18 (19 r = x i + y j = 1 2 at 2 + v 0 t + r 0 (20 ( j t 2 + (v 0 cos θ i + v0 sin θ j t (21 = 1 2 4

5 Rearranin, we et: r = (v 0 cos θ t i + ( 1 2 t2 + v 0 sin θ t j (22 Therefore, in terms of the scalar components: r { x = v 0 cos θ t y = 1 2 t2 + v 0 sin θ t (23 Applyin v = d r dt, we obtain the followin scalar components of the velocity vector: v { v x = v 0 cos θ v y = t + v 0 sin θ (24 Note that this motion can be considered as a sum of two motions, one alon the x-axis, which is uniform rectilinear as it appears from the expression of v x which is constant, and the other is alon the y-axis, which is uniformly accelerated, as it appears from the expression of v y since its derivative which is the acceleration alon y-direction is which is constant Time needed to reach the hihest point in the trajectory At the hihest point of the trajectory, the object stops movin up, and then starts to fall down. So the y-component of the velocity v y, which is responsible of the vertical motion of the object, will be zero at this moment: v y = t M + v 0 sin θ = 0 (25 where t M is the time taken by the projectile to reach the maximum point on the trajectory. Solvin equation 25 for t M, we et: t M = v 0 sin θ ( Maximum heiht The maximum heiht that the projectile reach can be found by substitutin t M in the equation of expression of y (that is, substitutin equation 26 in equation 23: y M = 1 2 t2 M + v 0 sin θ t M (27 where y M is the maximum heiht that the projectile reaches at the instant t M. Substitutin for the value of t M in equation 27, we et: 5

6 ( y M = 1 2 v0 sin θ 2 + v 0 sin θ = 1 2 v 2 0 sin2 θ ( v0 sin θ ( Equation of the trajectory (elimination of t: x = v 0 cos θ t (29 t = x v 0 cos θ (30 Substitutin equation 30 in the expression of y (equation 23: y = 1 2 t2 + v 0 sin θ t ( x 2 ( x = v 0 sin θ v 0 cos θ v 0 cos θ y = 2v 2 0 cos2 θ x2 + tan θ x (31 This shows that y is of the form: which shows that the trajectory is parabolic Rane x R (Displacement on the x-axis: y = ax 2 + bx (32 The distance covered on the x-axis, by the object can be found by calculatin the abscissa of the point of intersection of the trajectory with the x-axis. At the intersection of the trajectory with the x-axis, y = 0: y = t2 + v 0 sin θ t = 0 ( 1 2 t + v 0 sin θ t = 0 (33 Therefore there are two solutions of this equation: 1- t = 0, which represents the time at which the projectile was just launched. This solution when substituted in the expression of x, it ives x = 0, which represents the first point of intersection of the trajectory with the x-axis. We are not interested in this point; we are interested in the second point of intersection with the x-axis, which ives us the rane t + v 0 sin θ = t = v 0 sin θ t = 2v 0 sin θ, which is the time taken by the object to return to the zero-heiht. Substitutin in the expression of x (equation 23: 6

7 ( 2v0 sin θ x = v 0 cos θ t = v 0 cos θ Applyin the trionometric formula: = v2 0 (2 cos θ sin θ (34 Therefore: 2 cos θ sin θ = sin (2θ (35 x = x R = v2 0 (2 cos θ sin θ = v2 0 sin (2θ x R = v2 0 sin (2θ (36 Note that the same rane is obtained for two complementary anles. For example, the launchin anles 30 and 60 ive the same rane, which means both anles makes the projectile reach to the same point on the horizontal axis. Physics Zone by Farid Minawi 7