First and foremost, always begin the solution to your projectile motion problem by listing your given information. You should also make a sketch of your particular projectile's situation, if possible. Sketches assist in conceptualizing your particular problem, enhance the value and quality of your given information, and often provide useful data and promote ideas not provided in the problem text. Keep the following "projectile facts" in mind when creating and interpreting your sketch:
1. All projectiles travel in upward pointing, symmetrical, parabolic paths.
2. The apex (tip) of the projectile represents the highest point in the projectiles path, that is, the projectile's highest highest altitude is achieved here. It is also the exact mid-point of the projectile's total trip.
3. Our projectiles always take-off from and land on the same horizontal surface.
4. Air resistance as well as other external influences on the projectile's motion are neglected.
5. The angle of take-off of a projectile always equals its angle of impact.
6. A projectile's path is bilaterally symmetrical. This means that if we draw a line vertically downward from the apex to the ground line, we divide the trajectory into two equal segments, with each being the "mirror image" of the other.
7. The vertical line which bisects the projectile's trajectory separates it into "rising" and "falling" segments.
8. A projectile's motion consists of two concurrent yet independent components, one of which is oriented in the horizontal direction and the other which is oriented in the horizontal direction.
10. The horizontal component of motion is governed and described by constant rate motion equations.
11. The vertical component of motion is governed and described by ULAM (free-fall) equations in which acceleration is always directed vertically downward and is equal to 9.80 m/sexp2.
12. Gravity is always oppositional to the initial vertical component of motion of a projectile.
13. The "lifetime" of a projectile is always equal to the time it takes to fall back to the ground from take-off and is calculated from ULAM (free-fall) equations.
14. The time it takes for a projectile to reach its highest point (highest altitude; apex) is the same amount of time it takes for the projectile to return to the ground from its highest point.
15. The horizontal velocity component of a projectile's motion remains the same throughout its travel.
16. The vertical velocity component of a projectile's motion is always changing. It attains its highest value at take-off and impact, whereas it equals zero at the mid-point (apex; highest altitude).
17. The vertical velocity components at take-off and impact possess the same magnitude but opposite direction.
18. The horizontal distance traveled by a projectile is called its range.
Projectile Motion Problem-Solving Steps
1. Always resolve the projectile's velocity vector into its vertical and horizontal velocity components first. These are obtained from the following equations:
vx = v(cosA) and viy = v(sinA)
where A is the angle of inclination, v is the initial velocity of the projectile (m/s), vx is the horizontal velocity component (m/s), and viy is the initial vertical velocity component (m/s).
2. Next, calculate the time it takes for the projectile to reach the mid-point. This is determined from the following equation:
tm = viy/g
where tm is the time to reach the mid-point (s), vy is the vertical velocity component (m/s), and g is 9.80 m/sexp2.
3. Next, calculate the total time of the projectile's motion by doubling tm, that is
2tm = tt
where tm is the time to reach the mid-point (s) and tt is the total time of flight (s).
4. The range of the projectile is determined from the following equation:
dx = vxtt
where dx is the range of the projectile (m), vx is the horizontal velocity component (m/s), and tt is the total time of flight (s).
5. The altitude of a projectile at any time during its flight is calculated from the following equation:
dy = viyt - 1/2(gtexp2)
where dy is the altitude of the projectile (m), vy is the horizontal velocity component (m/s), g is 9.80 m/sexp2, and t is the time of interest (s). (Note that the maximum time a projectile can possess is tt.)
6. The maximum altitude of a projectile is determined from the following equation:
dymax = viyt - 1/2(gtmexp2)
where dymax is the maximum altitude of the projectile (m), vy is the horizontal velocity component (m/s), g is 9.80 m/sexp2, and tm is the time to reach the midpoint (s).
7. Some projectile problems involve objects thrown horizontally from a higher surface to a lower surface. Since the projectile only experiences forward and falling motion and possesses an initial vertical velocity component equal to 0 m/s, these can be solved as simple free-fall problems. As with other free-fall problems, the altitude, final vertical velocity component, and time to fall from the thrower's hand are calculated from (one or more) of the following equations:
tf = vy/g
vyexp2 = 2gdy
dy = 1/2(gtfexp2)
where g is 9.80 m/sexp2, vy is final vertical velocity component (m/s), dy is vertical distance traveled (m) and tf is the time of fall (s).