Question

A small child gives a plastic frog a big push at the bottom of a slippery 2.0-m-long, 1.0-m-high ramp, starting it with a speed of 5.0 m/s. What is the frog’s speed as it flies off the top of the ramp?

Short answer

The speed of the frog at the top of the ramp is 2.2 m/sec

Step-by-step solution

Step 1. The slant distance of the ramp is S= 2m

The height of the ramp is $H=1m$
The speed of the frog is $u=5m/sec$

Step 2. To determine the final speed of the frog

The acceleration of the frog is given by $a=g\sin \left(Ө\right)$
Here, localid="1648212517867" $\sin \left(ϴ\right)=\frac{Height\left(H\right)}{Slantheight\left(S\right)}$

Substitute the values of slant height (S) and height (H)
$\sin ϴ=\frac{1}{2}$

The acceleration due to gravity is $g=10m/s^2$
Thus, the acceleration of the frog is localid="1648212652084" $a=10\left(\frac{1}{2}\right)=5m/s^2$

Now, write the equation of motion,

$v^2-u^2=2aS$
substitute the known values,

$$\begin{gathered} v^2-5^2=2(-5)\left(2\right) \\ {v}^2-25=-20 \\ {v}^2=5 \\ \Rightarrow v=\sqrt{5}m/sec \end{gathered}$$
Thus, the speed of the frog at the top of the ramp is 2.2 m/sec