Check out some similar questions!

gigglez3576 · May 4, 2011, 01:02 PM

A stone thrown at an angle in such a way that the horizontal component of its initial velocity is 20 m/s and the vertical component of its initial velocity is 40 m/s. How far does the stone travel before landing? How high does the stone travel before landing?

When I calculated the time I got 4.08 s
From that I got that the distance traveled in the x direction was 81.63
For the height I got 163.27 Did I do this correctly?

ebaines · May 4, 2011, 02:09 PM

Originally Posted by gigglez3576

When i calculated the time i got 4.08 s

Not quite - that's the time for the stone to reach its maximum height. It will take another 4.08 second to fall back to the ground. This comes from this formula:

$ d = d_0 + v_0 t + \frac 1 2 a t^2 $

where
$ d = 0\\ d_0 = 0\\ v_0 = 40 \frac m s\\ a = -9.8 \frac m {s^2} $

and solve for t.

Originally Posted by gigglez3576

From that i got that the distance traveled in the x direction was 81.63

No, you need to fix the value for total time.

Originally Posted by gigglez3576

For the height i got 163.27 Did i do this correctly?

No - not sure how you got that result. The max height can be found in a couple of ways. You can use

$ d = d_0 + v_0 t + \frac 1 2 a t^2 $

and set $d_0 = 0, v_0 = 40$ m/s, and $t = 4.08$ s. Or you can use

$ v_2^2 - v_1 ^2 = 2ad $
where $v_2 = 0 ,\ v_1 = 40$ m/s, and solve for $d$ .

gigglez3576 · May 5, 2011, 10:12 AM

Ok so I redid the problem and I got t equal to 9.79 seconds is that correct?

Unknown008 · May 5, 2011, 10:32 AM

No, could you show how you got that time please?