AN INTRODUCTION TO NEWTONIAN MECHANICS by Edward Kluk Dickinson State University, Dickinson ND

SLANT PROJECTILE MOTION

        A few words about vectors
       Vectors are mathematical objects with two attributes, magnitude and direction. Take a speed, it has a magnitude telling how fast something with this speed is moving. But there is no information about direction of this motion there. Consequently a speed is not a vector. Talking about horizontal projectile motion we were using vectors without naming them. As a matter of fact, instead using there terms "horizontal component of speed" and "vertical component of speed" we should be using  horizontal and vertical components of velocity. Velocity is a vector showing both speed (which is its magnitude) and direction of motion at a given instant. It is intuitively clear that the velocity must be at every point along (tangential to) the path of the motion. If you are a vector wizard it is clear for you that any horizontal projectile motion is a vectorial superposition (or a vectorial sum) of two motions, horizontal uniform motion and vertical free fall. If you are not, but you still understand how horizontal projectile motion is composed of these two motions, it is good enough.

       A vertical projectile motion
       Let us introduce the same two dimensional Cartesian frame of reference as before, with an origin at zero height and zero displacement which is in the lower left corner of the vision field,  x axis directed horizontally to the right and y axis directed vertically up. Set the height of the object in the applet to zero, its initial speed  v_o to 0.45 m/s and the projection angle to +90^o. If there were no gravity acceleration the  object should be moving up with the constant speed equal to  v_o and its y coordinate should change proportionally with time according to the formula  y(t) = v_ot . In the presence of gravity acceleration g, but with the initial speed equal to zero and no support the object should be falling down with  y(t) = - (g/2)t². Thus, with the gravity acceleration g and the initial velocity v_o we may expect to see a simple combination of these two motions with

 y(t) = v_ot - (g/2)t²

and velocity

v(t) = v_o - gt .

According to the last formula the motion should start upward  (t = 0) with velocity 0.45 m/s . This velocity due to gravity acceleration should gradually decrease reaching zero at the time  t_mh = vo / g = 45 s because on the surface of this strange planet g = 0.01 m / s² . At this particular instant our object will reach its maximal height y_mh . We can express this height with help of  v_o  and  g  inserting  t = t_mh = vo / g  into formula for y(t) . A simple algebra leads to the following result  y_mh = v_o² / (2g) = 10.1 m . Right after the object reaches its maximal height the velocity v(t) becomes increasingly negative which means that it falls faster and faster down. Finally at a particular time instant t_f it will reach its starting point with y = 0. Inserting the total time of flight  t_f  into our expression for y(t) we obtain  0 = v_ot_f - (g/2)t_f². Solving it for t_f  we obtain t_f = 2vo / g which is twice of   t_mh . This means that the flight up takes the same amount of time as the fall down. The final velocity  v_f  at the instance when the object hits the ground equals to v(t_f) = - v_o .

       Now find "experimentally" y_mh . To make it easier set the tracer on and start the motion. Notice that initially  y = 0  for the bottom of the object. Consequently you must find highest position of the bottom of the object and compare it with the theoretical prediction for y_mh . Try to remember what height in the vision field was reached by the top of the object. It will help you to measure more precisely t_mh . Difficulties with precise measurements of  t_mh arise because close to its maximum height the object moves so slowly that it is not ease to catch the turning point. For next two runs switch the tracer off, measure t_mh and  t_f , and compare them with theoretical predictions.

    A slant projectile motion
    If the initial velocity of the object creates an angle ß with the horizontal, then beside of a vertical component the object motion also has a horizontal component. We may try to assume that each of these components depends only on a proper component of the initial velocity. This means, the vertical component of the motion depends only on vertical component of the initial velocity whereas horizontal component of the motion depends only on horizontal component of the initial velocity. Being a vector wizard you know what does it mean. If, however, you are not familiar with vectors you have a problem with understanding what exactly are these vertical and horizontal components of velocity. You certainly understand that they are directed respectively vertically and horizontally, but what about their magnitudes? In times of Galileo and Newton people who investigated this kind of motion had the same problem. A formal theory of vectors was created later. A possible solution of our problem is shown in the first figure. An initial velocity  v_in  which creates an

angle ß with the horizontal direction is resolved there onto two components v_inx and v_iny with help of perpendicular projections on x and y axes. If such model of resolution will work only experiments can answer.

       Expanding our theory we may assume that the vertical component of a slant projectile motion is governed only by v_iny . Consequently the vertical coordinate of the object

 y(t) = v_inyt - (g/2)t²

and its vertical component of velocity

v_y(t) = v_iny - gt .

Similarly for horizontal coordinate of the object

 x(t) = v_inxt

and its horizontal component of velocity

v_x(t) = v_inx .

The last two expressions differ from the former two by lack of the gravity acceleration factor because this acceleration occurs only in the vertical direction. As a matter of fact the vertical direction is defined by the direction of gravity acceleration.

       Since we have assumed the vertical and horizontal components of  slant projectile motion independent, then all results obtained for vertical projectile motion are valid for the vertical component of slant projectile motion. Only thing we have to change there is to replace  v_o  by v_iny . Thus  t_mh = v_iny / g ,  y_mh = v_iny² / (2g) and t_f = 2v_iny / g . Another interesting magnitude related to the horizontal component of the motion is the range of the projection x_r = x(t_f) = 2v_inxv_iny /g .

       Making our "experiments" with projectile motion we control magnitude of initial velocity (or speed) and angle of projection. In the theoretical formulae which were derived above we operate with v_inx and v_iny . If you know basic trigonometry and look at the first figure above the following relations become obvious

v_inx = v_in cos ß        v_iny = v_in sin ß .

If you do not, just use a calculator with cos and sin functions to find both components of initial velocity. It is also possible to solve this problem graphically. Draw a large scale graph like in the first figure above with a required angle ß assigning to a magnitude of required initial velocity a length of 10 cm. Find how many m/s represents 1 cm on your drawing. Measure components of velocity in cm from the drawing and recalculate them in m/s .

       Preparing an "experiment" with projection angles 30^o, 45^o and 60^o and a magnitude of initial velocity of  0.30 m/s you should obtain the following values for velocity components and theoretical predictions for t_f , x_r and y_mh

Now make this "experiment" setting the tracer on and compare the theoretical predictions with your "experimental" results. If they are reasonably close (up to 5% of difference) then the theory constructed above, including the method of resolving velocity onto the vertical and horizontal components, works.

      Analysing both, theoretical and experimental data for x_r you probably already have noticed that x_r for 30^o and 60^o is the same, and for 45^o  x_r  is the largest. Is it coincidence or rule?  Look at the second figure above and notice the following relations: if OB = OD then BF = DE, angle FOB = angle EOD and area OBFC = area OAED. If we assume that OF represents an initial velocity for projectile starting at the angle FOB from x axis, then OE must represent an initial velocity for another projectile starting with the same speed and at the same angle FOB from y axis. Moreover, the area OBFC for the first projectile can be expressed as v_inxv_iny for this projectile. Similarly, the area OAED for the second projectile can be expressed as v_inxv_iny for this projectile. But these areas are equal. Therefore the products v_inxv_iny for both projectiles must be also equal in spite of fact that (with exception of  angle FOB = angle EOD =  45^{o )} these projectiles have different x components of  their initial velocities and different y components of  their initial velocities. Consequently  x_r  for both projectiles is the same because x_r = 2v_inxv_iny /g . Therefore if two projectiles start with the same speed, the first at angle ß with respect to a horizontal axis and the second at the angle ß with respect to vertical axis, their ranges are the same. Now it should be intuitively clear why for a fixed speed the projection angle of  45^o gives the longest range. A formal proof of it is rather simple. From Pythagorean theorem we obtain

v_in² = v_inx² + v_iny².

It is also ease to verify the following identity

2v_inxv_iny = v_in² - (v_inx - v_iny)².

For a fixed speed of a projectile v_in² does not depend on the projectile angle. Therefore the left hand side of the identity reaches maximal value if  v_inx = v_iny which occurs if the projection angle is equal to 45^o.

       Evaluation
        If at this point you can solve the following problems:

• A first object has the projection angle 60^o and initial speed 0.35 m/s. A second object has the projection angle 30^o and initial speed 0.30 m/s. Which of them will reach a longer range? Try to solve this problem without doing any calculations.
• Generalize the formula for y(t) coordinate for a slant projectile motion if it starts from the level y_o instead of the level zero.
• Predict how far the object in the vision field will drop if it starts from the level of  5 m with initial speed of  0.35 m/s at the angle of 45^o  below the horizontal line (negative angle).
• Predict how high the object in the field of view will stop if you start it from zero level with the speed 0.45 m/s at the angle of 70^o. Notice that the object stops if it reaches the edge of the field of view and this stop is abrupt. What will be its horizontal and vertical component of velocity just before it stops. One of them will be negative!!

the objectives of this lesson are fully achieved. If you have doubts try to read it once more concentrating on them, but do not try to memorize this text. Physics is not about memorizing, it is about understanding.