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April 24, 2014

April 24, 2014

Posted by **Megan** on Friday, May 25, 2012 at 2:42pm.

Can someone please explain the correct equation to use thoroughly and help me to find the correct answer?!

- Physics -
**Steve**, Friday, May 25, 2012 at 2:52pmby speed, do you mean the vector sum of the horizontal and vertical speeds?

Must be, since the horizontal speed never changes, and the initial vertical speed was 0, so double that is still 0.

**v**= 4.3**i**- 9.8t**j**

|**v**| = √(4.3^2 + (9.8t)^2) = √(96.04t^2 + 18.49)

so, double the original speed is 8.6m/s and we want to find t when

8.6 = √(96.04t^2 + 18.49)

t = 0.76 sec

- Physics -
**Elena**, Friday, May 25, 2012 at 2:54pmThis is the projectile thrown with horizontal velocity.

So

v(x) = const,

v(y) = g•t.

v ² = v(x)² +v(y)²

v =2•v(x),

[2v(v)]² = v(x)² +v(y)²,

3•v(x)² =v(y)² = (g•t)²,

t =v(x) •√3/g =4.3•1.73/9.8 =0.76 s.

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