- #1

- 33

- 0

I tried 29.4 + 8.25 m/s(x) + 1/2(-9.8m/s)(x), but that didn't turn out to be right. Can I get some help with this one?

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 motionman04
- Start date

- #1

- 33

- 0

I tried 29.4 + 8.25 m/s(x) + 1/2(-9.8m/s)(x), but that didn't turn out to be right. Can I get some help with this one?

- #2

Pyrrhus

Homework Helper

- 2,178

- 1

- #3

Doc Al

Mentor

- 45,089

- 1,396

I assume you are trying to apply the following kinematic equation:motionman04 said:I tried 29.4 + 8.25 m/s(x) + 1/2(-9.8m/s)(x), but that didn't turn out to be right. Can I get some help with this one?

y = y_0 + v_0 t + (1/2)a t^2

Be sure to use a consistent sign convention: not only is the acceleration negative (a = - 9.8 m/s^2), don't forget that the initial velocity is also negative since it is thrown

And, as Cyclovenom points out, the final postion is where y = 0. Solve for t.

Last edited:

- #4

Integral

Staff Emeritus

Science Advisor

Gold Member

- 7,201

- 56

29.4 + 8.25 m/s(x) +1/2(-9.8m/s)(x),

Don't forget that the x (I would prefer t!) in the bold quantity needs to be squared.

Share: