- #1

- 9

- 0

**above/below**where it was launched?

Thanks

- Thread starter MattTuc13
- Start date

- #1

- 9

- 0

Thanks

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

Of course. Just use the same equations you would if the landing point were on the same level:

x= v_{0}cos(θ)t, y= v_{0}sin(θ)t- (g/2)t^{2}.

Let y_{0} be the height of the landing point relative to the starting point (0 is same level, negative if below, positive if above). Solve

y= v_{0}sin(θ)t- (g/2)t^{2}= y_{0} for t. (You will get two solutions the time for the landing is the larger of the two).

Since that is a quadratic equation, use the quadratic formula:

[tex]t= \frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}[/tex]

Now put that value of t in the equation for x (use the positive sign for the root to get the larger of the two solutions):

[tex]x= v_0cos(\theta)\frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}[/tex]

A bit more complicated than the case y_{0}= 0 but still a valid formula. Notice that when y_{0}= 0 that reduces to

[tex]x= \frac{2v_0sin(\theta)cos(\theta)}{g}[/tex]

the usual formula.

I would consider it easier to solve the equations than to memorize that formula.

x= v

Let y

y= v

Since that is a quadratic equation, use the quadratic formula:

[tex]t= \frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}[/tex]

Now put that value of t in the equation for x (use the positive sign for the root to get the larger of the two solutions):

[tex]x= v_0cos(\theta)\frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}[/tex]

A bit more complicated than the case y

[tex]x= \frac{2v_0sin(\theta)cos(\theta)}{g}[/tex]

the usual formula.

I would consider it easier to solve the equations than to memorize that formula.

Last edited by a moderator:

- #3

Päällikkö

Homework Helper

- 519

- 11

The last equation (where y_{0} = 0) should be

[tex]x= \frac{2v_0^2sin(\theta)cos(\theta)}{g}[/tex]

(notice the v_{0}^{2})

Which simplifies to:

[tex]x= \frac{v_0^2sin(2 \theta)}{g}[/tex]

[tex]x= \frac{2v_0^2sin(\theta)cos(\theta)}{g}[/tex]

(notice the v

Which simplifies to:

[tex]x= \frac{v_0^2sin(2 \theta)}{g}[/tex]

Last edited:

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 0

- Views
- 956

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 5K

- Replies
- 8

- Views
- 11K

- Replies
- 23

- Views
- 2K

- Replies
- 3

- Views
- 703

- Replies
- 2

- Views
- 969

- Replies
- 8

- Views
- 1K