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

Sam tosses a ball horizontally off a footbridge at 4.3 m/s. How much time passes after he releases it until its speed doubles?

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 2:52pm
by 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 2:54pm
This 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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