# Classical mechanics: Projectile Motion

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1. Nov 24, 2015

### REVIANNA

1. The problem statement, all variables and given/known data
Two projectiles are launched simultaneously from the same point above the flat terrain. The initial speeds of the projectiles are the same. Each projectile’s velocity makes the same angle with the horizontal. However, projectile A is launched above the horizontal and projectile B is below the horizontal.

Consider the entire flights of the projectiles. State which of the projectiles has greater:

1-time in the air

It's A
because
x_f = x_i+v sin(theta) t_f-(0.5*g*(t_f^2))

for A : 0=x_i + v sin(theta) t_f-(0.5*g*(t_f^2))

so t_f =sqrt ((2 v sin(theta)+x_y)/g)

for B : 0=x_i - v sin(theta) t_f-(0.5*g*(t_f^2))

so t_f =sqrt ((2 v sin(theta)-x_y)/g)

and to put it more simply the simulation picture shows it.

2-horizontal range
since the horizontal velocity vector is constant and equal for both A and B
and time in the air is more for A
A's horizontal range is obviously larger.

3-Landing speed
this is what I thought -
B is thrown with a downward vertical velocity so it would get increased as the acceleration due to gravity is also downwards
and A is thrown with a upward vertical velocity so it would get decreased as the acceleration due to gravity is downwards
and the horizontal velocities are equal so the landing speed of B would be greater.

but the answer says that the landing speed is same for both A and B.
why is it this way? what am I missing?

Last edited: Nov 24, 2015
2. Nov 24, 2015

### .Scott

Notice that there is a horizontal line extending to the right from the cannon. If you extend that line further to the right, it will cross the path of A - call that intersection point "w".
What do you suppose the velocity of A is when it crosses point w? Where do you suppose path "A" peaks relative to the cannon and point "w"? Since A is following a parabolic path, how do you suppose the and of Path A at the gun and the angle of path A at point "w" are related?

There are other ways to see this. If you launch an ball into the air so that it is traveling at 50mph as it leaves you hand 5 feet from the ground, how fast do you think it will be traveling downward when you try to catch it again - with your hand 5 feet above the ground? Assuming no losses to air friction. If it is anything other than 50MPH, where would the extra energy have come from or gone to?

3. Nov 24, 2015

### REVIANNA

at w:
velocity will be -v*sin(theta)
and angle will simply be -theta

at the point where A strikes the ground
vertical velocity= - v*sin(theta)- gt
and angle will be :
arctan((-vsin(theta)-gt)/vcos(theta))

I don't understand what you mean.

4. Nov 24, 2015

### .Scott

This much is all you needed. If you get this, then you should realize that the remainder of the A path is exactly like the B path.
I was hoping that you would recognize that the parabolic shape was left-right symmetric between the cannon and point w. It we take the cannon to be at 0, point w/2 would be the midpoint between the cannon and w, so it would be the peak location of the A.

5. Nov 24, 2015

### REVIANNA

I don't know why I could not see that
What I was missing was that they have the same trajectory once A crosses pt w and the extra time and longer horizontal range is just because of the path before it crosses W.
And yes its symmetric about the peak that's why it takes the same time to reach the bottom (W) as it took to reach the peak

and because they have the same initial velocity and theta and g they have the same LANDING SPEED.
W is an imp pt!

thank you