- #1

- 4

- 0

For instance, lets 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 lets 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 lets 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 lets 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**

V = 5 m/s

(4) Now lets 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 :uhh: 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?