Let us define projectile motion as the motion of a particle through a region of space where it is subject to constant acceleration. An object moving through the air near the surface of the earth is subject to the constant gravitational acceleration g, directed downward. If no other forces are acting on the object, i.e. if the object does not have a propulsion system and we neglect air resistance, then the motion of the object is projectile motion.
Assume that we want to describe the motion of such an object, starting at time t = 0. Let us orient our coordinate system such that one of the axes, say the y-axis, points upward. Then a = ayj =- gj, ay = -g. We can rotate our coordinate system about the y-axis until the velocity vector of the object at t = 0 lies in the x-y plane, and we can choose the origin of our coordinate system to be at the position of the object at t = 0.
Since we have motion with constant acceleration, the velocity at time t will be v = v0 + at, and the position will be r = r0 + v0t + (1/2)at2. Writing this equation in component form and using r0 = 0, v0 = v0xi + v0yj, and a = ayj = -gj we have
v = v0xi + v0yj - gtj, r = v0xti + v0ytj - (1/2)gt2j.
The position vector r and the velocity vector v at time t have only components along the x- and y-axes. By choosing a convenient orientation of our coordinate system we have simplified the mathematics involved in solving a projectile motion problem. We can treat projectile motion as motion in two dimensions with
vx = v0x, x = v0xt,
vy= v0y - gt, y = v0yt - (1/2)gt2.
If the initial velocity v0 makes an angle q0 with the x-axis, then v0x = v0cosq0 and v0y = v0sinq0.
Then
vx = v0cosq0 = constant, x = v0cosq0t,
vy = v0sinq0 - gt, y = v0sinq0t - (1/2)gt2.
We can solve x = v0cosq0t for t in terms of x, t = x/(v0cosq0) and substitute this expression for t into y = v0sinq0t - (1/2)gt2. We obtain
y = xtan(q0) - gx2/(2v02cos2(q0)),
an equation for the path (or trajectory) of the object. This equation is of the form y = ax - bx2, which is the equation of a parabola which passes through the origin. The trajectory for projectile motion is a parabola.
Links:
 | Horizontally launched projectile |
| Non-horizontally launched projectile |
Assume a projectile is launched with v0x = 4m/s, v0y = 3m/s. We have tanq0 = v0y/v0x = 3/4, q0 = 30.87o, v02 = v0x2 + v0y2, v0 = 5m/s. The projectile moves along a parabolic path until it impacts the ground. Its coordinates as a function of time are x = (4m/s)t, y = (3m/s)t - (4.9m/s2)t2. Its velocity components are vx = 4m/s, vy = (3m/s) - (9.8m/s2)t2.
| t | x(t) | y(t) | vx(t) | vy(t) |
0 |
0 |
0 |
4 |
3 |
0.05 |
0.2 |
0.13775 |
4 |
2.51 |
0.1 |
0.4 |
0.251 |
4 |
2.02 |
0.15 |
0.6 |
0.33975 |
4 |
1.53 |
0.2 |
0.8 |
0.404 |
4 |
1.04 |
0.25 |
1 |
0.44375 |
4 |
0.55 |
0.3 |
1.2 |
0.459 |
4 |
0.06 |
0.35 |
1.4 |
0.44975 |
4 |
-0.43 |
0.4 |
1.6 |
0.416 |
4 |
-0.92 |
0.45 |
1.8 |
0.35775 |
4 |
-1.41 |
0.5 |
2 |
0.275 |
4 |
-1.9 |
0.55 |
2.2 |
0.16775 |
4 |
-2.39 |
0.6 |
2.4 |
0.036 |
4 |
-2.88 |
As we can see from the graphs below, the projectile impacts the ground after approximately 0.6 seconds. It reaches its maximum height after approximately 0.3 seconds. Its range is approximately 2.4 meters. In approximately 0.3 seconds it has covered half its range. The trajectory for projectile motion is symmetric about the point of maximum height. The projectile covers the same horizontal distance reaching its maximum height as it does falling from its maximum height back to the ground. It takes the projectile the same amount of time reaching its maximum height as it does to fall from the maximum height back to the ground. When the projectile reaches its maximum height, after approximately 0.3s, the vertical component of its velocity is zero. We can therefore find the time when the projectile reaches its maximum height by setting vy = vy0 - gt = 0 and solving for t. We find tmax_height = vy0/g = v0sinq0/g. We can now find the range R by substituting t = 2tmax_height into the equation for x(t). R = v0cosq02tmax_height = (2v02cosq0sinq0)/g = (v02sin2q0)/g. Similarly, we find the maximum height h by substituting tmax_height into the equation for y(t).
h = v0sinq0tmax_height - (1/2)gt2max_height = (v02sin2q0)/2g.
Look at the expression for the range R, R = (v02sin2q0)/g. For a given v0, R as a function of the launch angle q0 has its maximum value when sin2q0 has its maximum value of 1. This happens when 2q0 = 90o, or q0 = 45o.
Link:
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Maximum Range
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Some problems involve the motion of two particles. Often these problems require that the two particles meet at the same place at the same time. An often stated problem is the monkey problem.
A monkey is hanging from a branch in a tree, a certain height h above the ground. A zookeeper stands a distance d from the tree, with a banana in his hand. The zookeeper knows that the monkey always lets go of the branch just as the banana is thrown to it. In which direction must the zookeeper throw the banana, so that the monkey can catch it?
Explore a simulation of this problem :
| Two Projectiles |
Analysis of the problem:
| The position of a projectile as a function of time is given by r = v0t + (1/2)gt2. We can view the motion of the projectile as a superposition of two motions, a motion with constant velocity v0 in the initial direction and a downward motion with constant acceleration, like the motion of a freely falling particle. If the gun is aimed at the target and fired at t = 0, then motion with constant velocity v0 will bring the projectile to the initial position of the target at some later time t. In the time interval between 0 and t the downward motion with constant acceleration carries the projectile downward by an amount (1/2)gt2. Superimposing the two motions will bring the projectile at time t to the position of the freely falling target at time t, independent of the magnitude of v0. In the time time interval between 0 and t the freely falling target moves downward by an amount (1/2) gt2. |
| One strategy in a snowball fight is to throw a snowball at a high angle
over level ground. While you opponent is watching the first one, you
throw a second snowball at a low angle timed to arrive before or at the same
time as the first one. Assume both snowballs are thrown with a speed
of 25m/s. The first is thrown at an angle of 70o with
respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first to arrive at the same time?
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| An astronaut on a strange planet finds that she can jump a maximum
horizontal distance of 15m if her initial speed is 3m/s. What is the
free-fall acceleration on the planet?
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| Two projectiles are thrown with the same magnitude of initial velocity,
one at an angle of q' with respect to the level
ground and the other at an angle 90o - q'.
Both projectiles will strike the ground at the same distance from the
projection point. Will both projectiles be in the air for the same
time interval?
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| A ball is projected horizontally from the top of a building. One
second later, another ball is projected horizontally from the same point,
with the same velocity. At what point in the motion will the balls be
closest to each other? Will the first ball always be traveling faster
than the second ball? What will be the time difference between when
the balls hit the ground? Can the horizontal projection velocity of
the second ball be changed, so that the balls arrive at the ground at the
same time?
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Links:
| Plane and Package |
| Truck and Ball |
| Virtual Cannon |
Links to other Web Materials
| Projectile Motion |
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