**Where are we going with this?**The information on this page relates to the skills needed to investigate and evaluate the graphical and mathematical relationship (using either manual graphing or computers) of one-dimensional kinematic parameters (distance, displacement, speed, velocity, acceleration) with respect to an object's position, direction of motion, and time.

**Projectile Motion**

**When Vertical Displacement Is Not Zero**

*Why is this even a thing!*

Source 2021-10-25 |

What happens if a projectile is set into motion, but from a position above the horizontal?

**Like in shot putting.**

The shot is launched from a distance of around 2.25 meters above the ground. How does this change the calculations?

The problem starts out just as if the projectile began on the ground.

The difference occurs in step 4 (below). Because the vertical displacement is NOT ZERO, the projectile has further to go before hitting the ground.

tdown is now NOT EQUAL to tup.

To find tdown, you will need to use the distance equation:

df = di + vit + 1/2at^{2}

Since we are calculating time from the point that the object stops moving up (and therefore is not moving at in the vertical direction) vi is zero and the middle term goes away. Hence, we can simplify the equation to:

df = di + 1/2at^{2}

Obviously, a is acceleration do to gravity, but what are di and df?

The most accurate way to think about this is to say

*df*is the ground and*di*is the total vertical displacement (*dy*) plus the initial vertical displacement (*dyi*). Since these would be measured from the ground to the top of the trajectory, they would be positive meaning a is in the opposite direction and is negative. Further*df*would be when it hits the ground and would be zero.Therefore:

df = di + 1/2at^{2}

where

di = dyi + dya = -9.81

Solving for t we find that

tdown = √ (dyi + dy) / 1/2(9.81)

So, if you are given theta and the velocity, here's a checklist sort of process…

- #1 Draw the diagram and label everything.
- #2 Find the component velocities in the x and y directions:

v • cosθ =vxv • sinθ =vy

- #3 Find the time up using
*vy*and acceleration due to gravity (probably 9.81 m/s/s)

tup =vy/ 9.81

- #4 Find the total time where…

t = tup + tdownand (different for dy ≠ zero as follows)

Find the maximum vertical displacement:

dy = vy•tup+ 1/2atup^{2}wherea = -9.81m/s/s,vywas found in step 2 andtupwas found in step 3.

df = di + 1/2at^{2}0 = (dyi + dy) + 1/2 (-9.81)t^{2}tdown = √ (dyi + dy) / 1/2(9.81)

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