Maximizing the Horizontal Distance of a Projectile

The well-known 45^o initial angle is for the case of the initial and final elevations being equal. This note will explain how to handle the general case. The range formula, x = (v_o²/g)sin2q is for the special case of the initial and final heights the same. It tells you the horizontal distance traveled (x) for a given initial velocity at angle q. This equation tells us that q = 45^o results in a maximum x, since sin2q is at its maximum value of 1 when 2q = 90^o. It will be shown (near the bottom of this page) that the range formula is the special case of x = (v_ov_f /g)sinf, where f is the angle between the initial and final velocities. Now for this equation, the initial and final elevations do not need to be the same. Note that this equation tells us that the maximum x occurs if the initial and final velocities are perpendicular. Skip down to An Easier Approach if you are in a hurry.

If the final height is above the initial, such as throwing snowballs uphill, the maximum x is achieved with q > 45^o, and if the initial height is above the final height, such as the shotput, q should be less than 45^o. We shall find the appropriate initial angle. It will be assumed here that one can muster the same v_o for any angle. This may not be exactly true, but it is very close if you exclude awkward cases like throwing something almost straight up.

Let y be the final height above the initial, so if the final is below, we make y<0.

The final v_y is -(v_yo² - 2gy)^1/2 , negative because it is on its way down.

Square v_y above, add it to v_x², and take the square root:

v_f=(v_x²+v_y²)^1/2 =(v_o²-2gy)^1/2 = final speed, the magnitude of the final velocity.

The time of flight is the change of v_y divided by -g, the acceleration in the y direction, and x = v_xt. From this, show

x = (v_ocosq/g)[( v_o² sin²q - 2gy)^1/2 + v_osinq]

A plot of x vs q will show that it has a maximum, and of course at the maximum x, the slope is zero.

So if you have had calculus, take the derivative of x with respect to q, set it equal to zero, and get this equation, with s and c representing sinq and cosq:

0 = -s(v_o²s²-2gy)^1/2 -v_os² + v_o²sc²/(v_o²s²-2gy)^1/2 + v_oc²

This is the conventional approach, but later we will present an easier method. After a lot of algebra you can show that

tanq = v_o/v_f (and sinq =v_o/(v_o²+v_f²)^1/2, cosq=v_f/(v_o²+v_f²)^1/2)

where q is the angle that will result in the maximum x. Deriving that will put more wrinkles on your brain. Do not attempt it unless you plan to spend hours on it.

You can check it by choosing y and v_o, calculate v_f, q and x, then recalculate x using larger and smaller angles.

Now for the maximum possible x in terms of v_o, g and v_f only, we use the expression for q in calculating x and after some algebra we find that

x_max = (v_ov_f /g)

An Easier Approach:

Play around with the geometry of q, v_o, and v_f required by the above equations, and you will see that the maximum x occurs when v_o and v_f vectors are perpendicular. This suggests that there might be a geometric method, and this turns out to be the case. The key to it is that when you vary the angles in a parallelogram, the maximum area occurs when the parallelogram is a rectangle.

Another key is that the linear motion equations have their corresponding vector equations, so
v_f = v_o + at

and the sketch below shows these vectors. Note that the dotted horizontal line is v_x, so that triangle has “area” = ¹/₂gtv_x. (area in units of v²) This is half of the parallelogram that has gt as a diagonal, and the tv_x part of the right hand side is the horizontal distance traveled, x. When the parallelogram is a rectangle its “area” is v_ov_f, so we find v_ov_f = gx_max, or
x_max = v_ov_f/g.

In general, if the angle between v_o and v_f is f, the area of the parallelogram is v_ov_f sinf, and we have
x =(v_ov_f sinf)/g.

Another useful equation is for average velocity: v_av = (v_o + v_f)/2 . Complete the parallelogram above, and the diagonal slanted toward the lower right is 2v_av. The diagonals of a rectangle are equal, so in the case of the maximum x, gt = 2v_av. Multiply both sides of this by t/2 and we find that
(1/2)gt² = v_avt. Now note that the right hand side, v_avt, is r, the slant distance from the starting point to the end point. (Who would have guessed that the slant distance r for the maximized case would turn out to be the same as the distance an object would fall in the same time? Fall distance from rest is (1/2)gt².)

r = v_ot + ¹/₂at² also, so the vectors above form an isosceles triangle when x = maximum.

The shotput: Muscle strength, body dimensions, and technique determine v_o and height of release. From the height of release and v_o we calculate v_f : (v_yf² = v_yo² - 2gy with y negative in this case. Add v_x² to both sides and get v_f² = v_o² - 2gy). Now calculate the angle above horizontal that one should strive for to maximize the distance: tanq = v_o/v_f.

Questions? fredrick.gram @ tri-c.edu (remove spaces)

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