Projectile Motion, different starting/ending height.

In summary, the conversation discusses different methods for solving projectile range problems. The quadratic equation is the most commonly used method, but the speaker proposes an alternative method involving finding the maximum height and adding the time for the projectile to fall back to its original height. However, this method is proven to be incorrect and the use of the quadratic equation is necessary to accurately calculate projectile range.
  • #1
TrueBlood
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If you are given the projectile angle, initial velocity, and height you can find the projectile range using the quadratic equation (solving for t). One thing I don't understand is why the quadratic equation seems to be the only way to solve these problems.

For instance, let's say an object is shot 8m/s, 40deg over the horizon, and at a height of 2.4 m. Now finding the range is straightforward.

(change in y) = Vot + 1/2at^2
-2.4 = Sin40*8*t + 1/2(-10)t^2
-2.4=5.12t + -5t^2
-5t^2+5.12t-2.4 = 0 This is the quadratic equation, which you solve for t


Now let's say you didn't want to use the quadratic equation.
(1) You know that the object travels 6.1 m in 1 sec.
Vf = Sin40*8 + (-10)t
0 = 5.12 -10t
tup = 0.5s
total projectile time (back to it's original starting height) = 1 s
.:. (change in x) = cos(40)*8*1s = 6.1 m
(2) Next let's find the velocity at the point, after 1 sec.
V(1 sec) = Sin(40)*8 + (-10)t
V(1 sec) = 5.12 -10
V(1 sec) = -4.88 This, of course, is the y-direction velocity only
(3) Now let's see what the final velocity is, at -2.4 m below it's starting point
V^2 = Vo^2 + 2a(change in y)
V^2 = (-4.88^2) + 2(-10)(-2.4) Vo is at the 1 second point, Vf is after that at a -2.4 height
V^2 = 23 - 48
V^2 = -25 Ok I get a negative number... no good, so I'll change it to a positive number even though that is mathematical sin :cool:
V = 5 m/s
(4) Now let's see how long it takes to get from -4.88 to -5m/s with gravity
Vf = Vo + at
-5 = -4.88 + (-10)t
t = 0.012
(5) Now add the calculated time (From 1 sec to end point) to the time it takes the projectile to get from starting point to 1 sec.
1 + 0.012 = TotalTime
(6) Now range is easy
(Change in x) = Cos40*8*1.012
(Change in x = 6.2 m

This of course is wrong :rolleyes: The actual time it adds to fall from 1sec to final-place is 0.372 sec (or 1.372 sec total). You get that with the quadratic equation... making the actual range 8.3 m.

So why doesn't my ingenious method of creating more work for myself not work?
 
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  • #2
Figured out a way to do it.

First find the max height , then the time it takes to fall from the max height to the ground (so height above original place + 2.4). Then add that time with the time it takes to reach the top. Then simply find range.
 

FAQ: Projectile Motion, different starting/ending height.

What is projectile motion?

Projectile motion refers to the motion of an object that is launched into the air and moves along a curved path under the influence of gravity.

What factors affect projectile motion?

The factors that affect projectile motion include the initial velocity, angle of launch, air resistance, and the acceleration due to gravity.

How does starting height affect projectile motion?

Starting height can affect projectile motion by changing the initial velocity and angle of launch, which can ultimately impact the trajectory and range of the object.

What is the difference between starting and ending height in projectile motion?

The starting height refers to the initial height of the object at the beginning of its motion, while the ending height refers to the final height of the object at the end of its motion. These heights can impact the trajectory and range of the object.

How can I calculate the trajectory of a projectile with different starting and ending heights?

The trajectory of a projectile with different starting and ending heights can be calculated using equations of motion, such as the kinematic equations, along with the known values of initial velocity, angle of launch, and acceleration due to gravity.

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