- #1

imhereyeah

- 3

- 0

I don't see how to find the gravitational acceleration unless the height off the ground from which the particle is shot is known, which it isn't.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter imhereyeah
- Start date

- #1

imhereyeah

- 3

- 0

I don't see how to find the gravitational acceleration unless the height off the ground from which the particle is shot is known, which it isn't.

- #2

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

y= v0 sin(theta)t- (g/2) t

In this problem we only know that v0= 10 m/s- theta is unknown.

In any case, the projectile reaches its "range" when y= 0 again: that means v0 sin(theta)t- (g/2) t

At that time x= vo cos(theta)(2 vo sin(theta))/g=

(2v0

We know that when v0= 10, the "maximum horizontal range" is 20 m.

Since the only variable is theta, differentiate the above with respect to theta to determine the value of theta that gives the maximum range. When you put that into the equation, the only "unknown" left is g so you can solve for that.

Once you know g, you can put that value, together with theta= 60 degrees in the above formula to find the range.

- #3

imhereyeah

- 3

- 0

I see what I was doing wrong, thank you.

Share:

- Replies
- 7

- Views
- 347

- Replies
- 2

- Views
- 358

- Last Post

- Replies
- 18

- Views
- 414

- Last Post

- Replies
- 6

- Views
- 288

- Replies
- 6

- Views
- 611

- Last Post

- Replies
- 20

- Views
- 621

- Last Post

- Replies
- 10

- Views
- 370

- Replies
- 3

- Views
- 335

- Replies
- 8

- Views
- 424

- Replies
- 35

- Views
- 754