## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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