A projectile is an object that moves through the air by virtue of its own inertia. Recall that mass is a measure of an object's inertia which is its resistance to a change in its state of motion. The term freefall means that the only force acting on the projectile is gravity; that is, there is no air resistance present. While in freefall, all projectiles experiences a unique value for their vertical acceleration: a = -g = -9.81 m/sec2. The term trajectory means the projectile?s path through the air. If the projectile only has vertical velocity, its trajectory traces out a vertical line. When it has a constant horizontal velocity combined with a vertical velocity which is uniformly accelerated, the trajectory will be parabolic. The term apex means highest point in the projectile?s trajectory where its instantaneous vertical velocity equals 0.
Refer to the following information for the next four questions.
Take a moment to remember the kinematics equations for uniformly accelerated motion.
What kinematics formula relates the variables s, vo, a and t ?
What kinematics formula relates vf, vo, a, and s?
What kinematics formula relates vf, vo, a, and t?
What kinematics formula relates s, vo , vf and t?
These formulas are used when the acceleration is uniform (constant). Remember that acceleration is the rate of change of velocity. If the object is either losing speed while traveling in a positive direction OR gaining speed in a negative direction, a is negative. In freefall problems, a has a value of -9.81 m/sec2.
This value is represented by the variable g. Which is either called the "acceleration due to gravity" or the "gravitational field strength." Its value depends on where the projectile is located with respect to the center of the earth. The value for g on the surface of the earth is derived based on the formula for universal gravitation and weight.
weight = force of universal gravitation |mg| = G(mME/r2) |g| = G(ME/r2)
Try it! G = 6.67 x 10-11 Nm2/kg2, ME = 5.98 x 1024 kg, r = RE = 6.37 x 106 m
Since gravity pulls objects towards the center of the earth, this acceleration has a downward direction, and g = -9.81 m/sec2.
For a projectile thrown vertically straight upwards, examine the sketch below which relates the graphs for the projectile's position vs time and its velocity vs time.
When doing freefall /projectile problems, vertical velocities, v, are
positive when the object is traveling "up" towards the apex, and
negative when the object is falling "down" after having reached the apex.
While the displacement , s, is
positive when the projectile is located "above the release height,"
negative when located "below the release height," and
equal to zero when the projectile has returned to its original release height.
Refer to the following information for the next five questions.
Identify the positions (A, B, C, D, or E) that represent each set of criteria.
vf > 0 and s > 0
vf < 0 and s < 0
vf < 0 and s > 0
vf = 0 and s > 0
vf < 0 and s = 0
Now let's apply our knowledge to some problems that contain numerical data. In each of the following scenarios, state the values of vo, a, and s.
A rock dropped from a 20 meter bridge falls into the river below.
A rock thrown upwards at 6 m/sec from a 20 meter bridge falls into the river below.
A rock thrown downwards at 6 m/sec from a 20 meter bridge falls into the river below.
PROJECTILE AND CHASE
spursuer = "gap" + sleader
vot + ½at2
number
vot + ½at2
vt
vt
Each column in the above table states the allowed behaviors for the pursuer and the leader. Each participant can either be experiencing accelerated or linear motion. The numerical value of the "gap" can be equal to zero (if the two objects start side-by-side) or it can be a nonzero number. The parameter t, for time, unites the equations. To solve chase equations, you first determine the time that is required for the two objects to come together - then, you use that time to determine the position of their collision.
To work this type of problem, one object is considered the leader and the other is the pursuer. The pursuer, in reaching the leader's final location, must not only close the leader's original gap but also account for any subsequent displacement the leader travels while being chased.
Example
A ball of mass 1.0 kg is dropped from rest from a height of 5.0 meters above the ground, as shown in the diagram at the right. It undergoes a perfectly elastic collision with the ground [that is, it leaves the surface of the ground with exactly the same amount of kinetic energy (KE = ½mv2) with which it initially strikes the ground] and rebounds at a speed of 9.9 m/sec. At the instant that the ball rebounds, a small blob of clay of mass 0.1 kg is released from rest from the original height H, directly above the ball, as shown. The clay blob while descending eventually collides with the ascending ball. Assume that air resistance is negligible.
To determine how much time passes after the clay blob is released until it strikes the ball we must first set up the chase equation. Let's assume that the 1.0-kg ball is the pursuer (since it is traveling in a positive direction) and that the 0.1-kg clay blob is the leader.
Next let's determine the height above the ground at which the collision takes place. Since this represents the distance that the ball traveled upwards after its bounce, we can substitute the time, 0.505 seconds, into the ball's equation.
s = 9.9t + ½(-9.8)t2 s = 9.9(0.505) + ½(-9.8)(0.505)2 s = 3.7 meters
The position-time graph for this problem would look like the one shown below. Note that the two parabolas intersect at a height of 3.7 meters and a time of 0.505 seconds. The purple parabola represents the clay blob which is gaining speed in a negative direction while the blue parabola represents the ball which is losing speed while traveling in a positive direction.
Next, let's determine how fast the ball and the clay were traveling at the moment that they collided.
clay (descending)
ball (ascending)
vf = vo + at
vf = vo + at
vf = 0 +(-9.8)(0.505)
vf = 9.9 + (-9.8)(0.505)
vf = -4.95 m/sec
vf = +4.95 m/sec
The velocity-time graph for this problem is shown below. Each line has the same slope of -9.8 m/sec2. The blue line starts at 9.9 m/sec and ends at 4.95 m/sec while the purple line starts are 0 m/sec and ends at -4.95 m/sec. Note that the blue area between the line and the x-axis must equal 3.7 meters while the purple area between the line and the x-axis must equal 5.0 - 3.7 or 1.3 meters. The combined blue and purple areas must equal the entire original gap, 5.0 meters.
Practice your skills with this type of problem by completing the accompanying worksheet entitled Chase Problems: Projectiles.
Hope it is useful to u
Whenever u feel bad go for math
if u feel too bad
imagine your rival competeing u
U will be energetic like never before
Moderation Team
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