#### Answer

The frog's speed at the top of the ramp is 2.3 m/s

#### Work Step by Step

We can find the angle of the ramp;
$sin(\theta) = \frac{1.0~m}{2.0~m}$
$\theta = arcsin(\frac{1.0~m}{2.0~m})$
$\theta = 30^{\circ}$
The acceleration directed down the incline is $g~sin(\theta)$. This means that;
$v^2 = v_0^2+2ad$
$v = \sqrt{v_0^2+2ad}$
$v = \sqrt{(5.0~m/s)^2+(2)(-9.80~m/s^2)~sin(30^{\circ})(2.0~m)}$
$v = 2.3~m/s$
The frog's speed at the top of the ramp is 2.3 m/s