What is a Projectile?
A projectile is any object that experiences horizontal and vertical motion simultaneously. The horizontal and vertical motion are independent of each other and are only linked by time. The path of a projectile is parabolic and is referred to as its trajectory.
Projectiles That are Not Launched
A projectile that is not launched is one that has no initial vertical velocity. Examples of projectiles like these are object that are launched strictly with a horizontal velocity, such as a stone that is thrown outward to an open field or body of water. An objects that rolls off a cliff such as a boulder or a pencil off a table is also considered a projectile.
The time that the projectile spends in the air is solely dependent upon the vertical height it falls. Specifically, the vertical height is directly related to the square of the time the object spends in the air. This time can then be used to find the horizontal range, which is the product of horizontal velocity and time.
How far will a ball land from a wall if the wall is 5.41 m high and it rolls off the wall with a velocity of 2.35 m/s?
To solve this problem, the time of flight must be determined. This can be found from the vertical height:
y = 0.5gt2
5.41 = 0.5(9.81)t2
1.10 = 2
t = 1.05 s
Now that the flight time for the projectile has been determined, the horizontal range can be found:
x = vt
x = (2.35)(1.05)
x = 2.47 m
Projectiles That are Launched: No Vertical Displacement
A projectile that is launched will have a non-zero vertical velocity. Any object that is launched at an angle will have a horizontal and vertical component to its velocity. In this first example, an object that returns to the same vertical displacement from which it was launched (e.g. a ball kicked into the air and returning to the ground farther down the field) will be examined.
A goalkeeper kicks a ball with a velocity of 18.9 m/s at an angle of 25 degrees to the pitch. How far downfield should an opposition player position himself to be where the ball lands?
First, the initial vertical velocity and horizontal velocity need to be determined. While the vertical component is immediately necessary, the horizontal component will be needed to solve this problem since it is about the range of the soccer ball.
uy = (u)(sinθ)
uy = (18.9)(sin 25)
uy = 7.99 m/s
ux = (u)(cosθ)
ux = (18.9)(cos 25)
ux = 17.1 m/s
Recall that the vertical displacement determines the time spent in air. Since the ball is assumed to be kicked off the ground and will return to the ground at the end of its path the vertical displacement is zero meters. In any situation where an object returns to the same position (vertically) from which it was launched the vertical displacement is zero. Substitutions for vertical displacement (sy) and initial vertical velocity (uy) can be made into the following equation:
sy = uyt + 0.5gt2
0 = 7.99t + 0.5(-9.81)t2
0 = t(7.99 - 4.91t)
Note that the value for g is negative because it acts in a downward direction, opposite the upward direction of the initial kick. Since the equation is a quadratic, two answers are possible for t. The last step shows how t can be factored out, leaving two solutions:
t = 0 and
7.99 - 4.91t = 0
The solutions for t are therefore 0 and 1.63 s. Both of these answers are reasonable since the ball is on the ground at the beginning (t = 0 s) and end (t = 1.63 s) of its journey. Since the objective of the question is to calculate range, the 1.63 s value is the one used to find range.
To find range, the initial horizontal velocity solved earlier (ux) will be used in the following equation:
sx = uxt
sx = (17.1)(1.63)
The range is 27.9 m.
Projectiles That are Launched: Maximum Height
Recall the scenario from the last problem. Consider a new question: What is the maximum height attained by the ball?
The initial vertical velocity uy is needed since the maximum height is dependent on the vertical component of the initial velocity. Its value, as found above is 7.99 m/s.
If the time of flight is known, this problem can be solved rather simply. Realize that at its maximum height, the vertical component of velocity is zero m/s. This is important for either method of solving this problem. In this first method, one-half the total time of flight will be used. In a scenario where the vertical displacement is zero, at one-half the total time of flight a projectile will be at its maximum height. Since the total time of flight is 1.63 s, one-half the time of flight is 0.815 s. This can be utilized in the following equation:
sy = uyt + 0.5gt2
sy = (7.99)(0.815) + 0.5(-9.81)(0.815)2
sy = 3.26 m
The first situation is not usually encountered as the time of flight may be unknown. Once again, by applying the fact that at its maximum height the vertical velocity will be zero, the height can be found using:
v2 = u2 + 2asy
02 = 7.992 + (2)(-9.81)sy
-63.8 = -19.6sy
sy = 3.26 m