# Proj20

The force of primary importance acting on a projectile is gravity. This is not to say that other forces do not exist, just that their effect is minimal in comparison. A tossed helium-filled balloon is not normally considered a projectile as the drag and buoyant forces on it are as significant as the weight. Helium-filled balloons can't be thrown long distances and don't normally fall. In contrast, a crashing airplane would be considered a projectile. Even though the drag and buoyant forces acting on it are much greater in absolute terms than they are on the balloon, gravity is what really drives a crashing airplane. The normal amounts of drag and buoyancy just aren't large enough to save the passengers on a doomed flight from an unfortunate end.

A projectile is any object with an initial non-zero, horizontal velocity whose acceleration is due to gravity alone.

taken from http://physics.info/projectiles/

Lets for example say a man has a tranquilizer gun trained on a monkey in the jungle, however this monkey has instaneous reflexs and lets go of the branch he is swinging on at precisely the same moment, how could the hunter amend his targeting?

And who says it has to be all work and no play

Play some Basketball Projectiles !!

http://library.thinkquest.org/16600/games/bball/

Heres another from Fowler. This one allows you to add in Air resistance, for a more actual appreciation of Rocket science

http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/ProjectileMotion/jarapplet.html

What if ..... said Newton and his mountain!

http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/newt/newtmtn.html

thanks fowler

The vertical and horizontal components of the motion are independent.

Projectiles

This topic is about objects which travel through the air under gravity. We analyse the problems by separating the x-direction (or i-direction) from the y-direction (or j-direction).

Within these two directions it is important to distinguish between two very different vectors: the first is the position vector, which describes where the particle is at any time; the other is the velocity vector which describes the motion of the particle wherever it is. We are not interested in how the particle came to be flying through the air, we just want to know where the particle is and at what velocity is it travelling as it moves.

There are four main types of problem:

1. Projectiles on a horizontal plane
2. Hitting a target
3. Projectiles on an inclined plane
4. Projectiles which bounce

Projectiles on a horizontal plane

If the particle is projected on a horizontal plane with initial speed u at an angle to the plane, then its velocity and displacement in each direction will be:    This next section is completely robbed from oliver murphys web site ...

• If you want to find the ‘time of flight’, find t when Sy = 0 or where the Sy = the distance that the object lands at (- for below the starting height)

• If you want to find when the particle reaches its highest point, find t when vy = 0

• If you want to find the range, find Sx when Sy = 0 .

time of flight

f you use this formula you must first derive it. Don’t produce it out of a hat!

• If you want to find the maximum height, find Sy when vy = 0 [The quickest way is to use the formula v2 = u2 + 2as ] The general equation for the maximum height is

.

highest point

f you use this formula you must first derive it. Don’t produce it out of a hat! • If you want to find the maximum height, find Sy when vy = 0 [The quickest way is to use the formula v2 = u2 + 2as ] The general equation for the maximum height is .

range

The general equation for the range is .

If you use this formula you must first derive it. Don’t produce it out of a hat!

• If you want to find the maximum height, find Sy when vy = 0 [The quickest way is to use the formula v2 = u2 + 2as ] The general equation for the maximum height is .

Again, show how to get this formula before you use it.

• A common question is to find the angle which gives maximum range and to find that range.

The answer is___________ which yields a range of

range = Sx @ Sy = 0

Sy = uyt - 1/2gt2 → uyt = 1/2 gt2 → t = uy/g = usinA / g

Sx = Uxt & t = usinA / g

Hitting a target

If the particle hits a target at the point (a , b), the you must solve the simultaneous equations

when . The method is always the same:

• Let • Substitute this expression for t in the second equation.

• Change and to get a quadratic equation in • Solve the quadratic equation!

Note, if the particle ‘just clears a wall’ the technique is just the same as if the particle hits a target at the top of the wall. Use the same steps as above.

A Walter special on Projectiles along a horizontal plane

Some very interesting questions.

http://physics.info/projectiles/practice.shtml

A link to a cannon applet

http://www.lon-capa.org/~mmp/kap3/cd060.htm

check this out for velocity vectors of a body falling with some horizontal velocity