By Ronald W. Leigh, Ph.D.
February 6, 2008
Copyright © 2008 Ronald W. Leigh
In this paper we discuss a few basic kinematics formulas, determine the launch angle that gives the greatest distance, and develop a formula for the range of a projectile given its initial velocity.
1 mph = 5280 feet / 3600 sec = 1.467 fps (Therefore fps is a slower unit than mph.)
If you know the number of mph, multiply by 1.467 to get fps.
If you know the number of fps, multiply by .6818 to get mph.
v = velocity (fps); d = distance (feet); t = time (sec);
a = acceleration (feet/sec²); ag = acceleration due to gravity (32.16 feet/sec² at the earth's surface)
|When velocity is constant:||When acceleration is constant:|
|(Equation 1) v = d / t
example: "miles per hour"
|(Equation 4) a = v / t
example: "0 to 60 in 5 sec."
|(Equation 2) d = v • t||(Equation 5) v = a • t|
|(Equation 3) t = d / v||(Equation 6) t = v / a|
The following diagram shows the parabolic path of the projectile, with the unknown angle A as the launch angle. It is assumed that the launch location and the landing location are at the same height. In this part of the calculation we keep V constant and vary A to find the launch angle that will give the greatest distance.
The parabolic path can be divided into two components. The vertical component is affected by gravity and thus experiences constant acceleration. The horizontal component is unaffected by gravity and thus experiences constant velocity.
The initial vertical velocity is V • sin(A)
The rising time can be calculated using equation 6.
t = v / a = V • sin(A) / a
The falling time is the same as the rising time, so the total air-time for the projectile is
Total time = 2V • sin(A) / a
The horizontal velocity is V • cos(A)
The horizontal distance can be calculated using equation 2.
d = v • t = V • cos(A) • 2V • sin(A) / a
In the above equation, V is constant, and a (gravity) is constant, so d is greatest when sin(A) • cos(A) is greatest, which can be discovered through several trial calculations to be 45°.
When the launch angle is 45°, both the initial vertical velocity and the horizontal velocity are V / √2
Regarding rising and falling time, using equation 6,
t = v / a = V / a√2
Total time = 2V / a√2
Regarding horizontal distance, using equation 2,
d = v • t = (V / √2) • (2V / a√2) = v² / a
Using the value for the acceleration of gravity, the formula for the range when launching at 45° becomes:
|d = v² / 32.16 feet|