Moving
What is the y component of the velocity vector?
This formula relates the horizontal displacement to horizontal velocity and time of flight.
What is
sx = vox t
value of initial vertical velocity (unless stated otherwise)
What is 0 m/s as long as we have a purely vertical launch
This motion remains constant during projectile motion.
What is the horizontal component of the velocity?
This formula relates the time to the vertical displacement and vertical acceleration.
What is
t= sqrt (2sy / ay)
we know that voy = 0 m/s and ay = g
so
we start with the no vf formula
sy = voyt + 1/2 ay t^2
sy = 1/2 ay t^2
t = sqrt (2sy/ay)
value of the acceleration on a horizontally launched object (as long as air resistance can be ignored)
What is 10 m/s^2 ?
This is responsible for changing the magnitude and direction of the projectile as it moves along its trajectory.
The only acceleration (as long as we can ignore air resistance is the acceleration due to the gravitational force which is -10 m/s^2 or -9.8 m/s^2
This formula relates horizontal displacement to vertical displacement, vertical acceleration, and initial velocity for the horizontal launch.
What is
sx = vox * sqrt (2sy/ay)
start with finding time in the y direction
t = sqrt (2sy/ay) using the no vf formula with voy = 0 m/s since we are dealing with a horizontal launch
now we know that sx = vox t
so we can substitute our value for t in
sx = vox (sqrt (2sy/ay))
the variable for this item controls the horizontal range of an object (the larger this is the larger our range - regardless of the fixed elevation)
what is vox or what is the initial horizontal velocity or what is vo for a horizontally launched object?
sy will determine time of flight the higher you are the longer your time of flight and indeed the larger your range for objects launched at the same initial speed.
However, the larger the speed the larger the horizontal range regardless of the elevation. keep the elevation the same and launch faster the object will move further horizontally.
If they want us to consider changes to both vo and sy
then we found that relationship to be
sx = vox * sqrt (2sy/ ay)
so in order for a smaller velocity to result in a larger horizontal displacement the
increase in vox has to be greater than the decrease you get from the reducing sy... which will change by the square root of the factor of change
so change in vox > sqrt of the change in sy in order for sx for a lower elevation to be farther than an object with a higher vox
For our launch problems, this value is zero at maximum height.
What is the y component of the velocity vector?
The only formula that can be used with horizontal variables for launch problems (unless there is air resistance).
What is vx =sx/t?
formula to calculate the final velocity of a horizontally launched object
vf = sqrt (vox^2 + vfy^2)
for the magnitude and
theta = inverse tan of (vy/vx)
then be careful about where this angle is
A projectile is dropped at the same time an identical projectile is launched horizontally. This will be true about the time it takes these to land, as long as they are launched from the same vertical height.
The time to land will be the same.
The vertical height influences the time of flight but since a horizontally launched object and a dropped object both have an initial y component of the velocity of 0 m/s and an acceleration of -10 m/s^2 they will cover the same distance in the same amount of time.