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**[SOLVED] Projectile Motion Problem**

## Homework Statement

A frog jumps from a log and lands on the ground a distance of 2.80 m away. The frog is airborne for 2.5 seconds, and leaves the log at an angle of 40 degrees. (a) What was the frog's speed (vo) just after it jumped? (b) How high above the ground did the frog jump from?

## Homework Equations

The equations I used:

[tex]

y = y_0 + v_0_yt - (1/2)g(t^2)

[/tex]

[tex]

x = x_0 + v_0_xt

[/tex]

## The Attempt at a Solution

First I found that

[tex]

v_0_y = v_0sin(40)

[/tex]

and

[tex]

v_0_x = v_0cos(40)

[/tex]

Then,

[tex]

x = x_0 + v_0_xt

[/tex]

[tex]

2.8 = 0 + (v_0*cos(40))(2.5)

[/tex]

[tex]

v_0 = 1.4621 m/s

[/tex]

Now,

[tex]

v_0_y = v_0sin(40)

[/tex]

[tex]

v_0_y = 1.4621sin(40)

[/tex]

[tex]

v_0_y = 0.9398 m/s

[/tex]

Using the first equation,

[tex]

y = y_0 + v_0_yt - (1/2)g(t^2)

[/tex]

[tex]

0 = y_0 + (0.9398)(2.5) - (1/2)9.8(2.5^2)

[/tex]

[tex]

y_0 = 28.28 m

[/tex]

...Now I really don't think that a frog would be jumping off a 30 meter log. Any ideas at where I'm going wrong?

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