A projectile is fired with an initial velocity equal to v₀ and at an angle θ₀ with the horizontal, the firing point and the target are on the same horizontal plane, and neglecting air resistance, determine:
a) The maximum height that the projectile reaches;
b) The time required to reach the maximum height;
c) The duration of the total movement;
d) The maximum horizontal range of the projectile;
e) The equation of the trajectory;
f) The angle of fire that provides the maximum range;
g) Show that shots with complementary angles have the same range;
h) The speed at any point on the trajectory;
i) The components of acceleration at any point on the trajectory.


Problem data:

• Initial speed:    v₀;
• Shooting angle with the horizontal:    θ₀.

Problem diagram:

We choose a reference frame with the axis Ox pointing to the right and Oy pointing in the upward direction, the acceleration due to gravity pointing downward, and the trigger point at the origin of the frame of reference (x₀, y₀) = (0, 0), (Figure 1).


Solution

The motion can be decomposed along the x and y axes, the components of the initial velocity v₀, in magnitude (Figure 2)
in uniform motion, v_x = v_0x is constant, we can substitute v_x by the value of (I) and S_0x = 0 In the y direction of the projectile, the acceleration due to gravity acts on it, the motion represented by a vertical launch during the ascent and a free fall on the descent, where S_0y = 0, v_0y is given by expression (II), and a = −g the acceleration due to gravity is the opposite direction of the reference frame. the velocity equation as a function of time is given by the equation of velocity as a function of displacement is given by In Figure 3, we see that in the motion along the x direction, we have that for equal time intervals we have equal displacement intervals (Δx₁ = Δx₂ = Δx₃ = Δx₄ = Δx₅ = Δx₆). In the y direction, we have that during the rising, for equal time intervals we have smaller displacement intervals, the projectile is being braked by the action of gravity (Δy₁ > Δy₂ > Δy₃) until the velocity v_y equals zero. During the fall, gravity begins to pull the projectile back to the ground with an accelerated velocity, for equal time intervals we have increasingly larger displacement intervals (Δy₄ < Δy₅ < Δy₆).


a) To find the maximum height, h_max, reached by the projectile, we analyze the motion along the y direction. When the projectile reaches maximum height its velocity v_y equals zero, v_y = 0, using the equation (VI)
b) The time of rising, t_R, to reach the maximum height will be obtained from equation (V) with the condition that the velocity equals zero at the maximum height reached by the projectile, v_y = 0
c) The total time, t_T, of the motion will be the sum of the rising time, t_R, and the descent time, t_D, in the vertical launch and free fall motion the rising and descent times are equal, we have the condition with t_R = t_D using the result for the rise time obtained in the previous item
d) The time calculated above for the projectile to rise and fall is also the time it will take to go from the origin to the point S_max along the x-axis, substituting the answer to the previous item in equation (III)
and substituting this identity in the equation above, we get the maximum range
e) To obtain the equation of motion, indicated in Figure 1, we have y as a function of x, or y = f(x), using equations (III) and (IV) for the movements in x and y and remembering that S_0x = S_0y = 0, we have the system solving the first equation for time substituting this value in the second equation Making the association shown below with a Quadratic Equation of the type y = ax²+bx+c we see that we obtained a function of the type S_y = f(S_x) with the coefficient a < 0 which indicates that our trajectory is a parabola concave downwards.

f) The answer obtained in item (d) for the maximum range, S_max, depends on the initial launch angle, from trigonometry we know that the sine function varies from −1 to 1, and the maximum range value occurs when


g) In trigonometry we have that complementary angles are those that add up to 90° or \( \frac{\pi }{2} \), two complementary angles θ₁ and θ₂
Writing the expression (VII) of θ₂ as a function of θ₁ and substituting in the equation (IX) we have \( \sin \pi =0 \), \( \cos \pi =-1 \) comparing equations (VIII) and (X)


In Figure 5-B we see that the vectors form a right triangle and the speed is calculated applying the Pythagorean Theorem using v_x = v_0x from expression (I), and v_y from expression (V) factoring \( v_{0}^{2} \) in the first two terms on the right-hand side of the equation and −2g in the third and fourth terms
in the first term in parentheses we apply the trigonometric identity above, the second term in parentheses can be obtained from equation (IV) of the position in the y direction with S₀ = S_0y if \( \Delta S_y=S_{y}-S_{0y} \), so the equation (XII) can be written
i) The acceleration due to gravity, \( \vec g \), to which the projectile is subjected at any point on the trajectory can be decomposed into the tangential acceleration, \( {\vec g}_t \), and the normal acceleration, \( {\vec g}_n \), which is perpendicular to the trajectory at the considered point (Figure 6-A). From Figure 6-C
substituting this cosine value in expression (XIII) the tangential acceleration will be In the same way as Figure 6-D and substituting in expression (XIV) for the normal acceleration