Question

A skateboarder starts up a 1.0-m-high, 30° ramp at a speed of 7.0 m/s. The skateboard wheels roll without friction. At the top she leaves the ramp and sails through the air. How far from the end of the ramp does the skateboarder touch down?

Short answer

The skateboarder touches down $\boxed{\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\mathit{m}}$from the end of the ramp.

Step-by-step solution

Given information

Given :

A skateboarder starts up : $1.0-m-$high, $30°$ramp

Speed of skateboarder :$7.0m/s$

Calculating the horizontal and vertical components of the initial velocity 

The problem can be divided into two halves.

First, the skater approaches the ramp, which is $30$degrees high and inclined. During this time, the skater's speed decreases owing to gravitational acceleration, and we can compute the skater's velocity at the end of the ramp.

${v_f}^2={v_i}^2+2a∆s$

Calculate the horizontal and vertical components of the initial velocity first.

Horizontal components are as follows:

role="math" localid="1650776976336" $$\begin{gathered} v_x=v_0\cos (30°) \\ =7.0\frac{m}{s}\times \cos (30°) \\ =\mathbf{}\mathbf{6}\mathbf{.}\mathbf{06}\mathbf{}\frac{\mathbf{m}}{\mathbf{s}} \end{gathered}$$

Components that are vertical:

$$\begin{gathered} v_x=v_0\sin (30°) \\ =7.0\frac{m}{s}\times \sin (30°) \\ =\mathbf{}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\frac{\mathbf{m}}{\mathbf{s}} \end{gathered}$$

Let's solve for "$v_{f_y}$" in the horizontal and vertical components of velocity as the skater leaves the ramp :

$$\begin{gathered} v_{f_y}²=v_y²-2g∆s \\ \Rightarrow v_{f_y}=\sqrt{\left|v_y²-2g∆s\right|} \\ =\sqrt{\left|3.5\frac{m²}{s}-2(9.80)\frac{m}{s^2}(1m)\right|} \\ =\mathbf{2}\mathbf{.}\mathbf{71}\frac{\mathbf{m}}{\mathbf{s}} \end{gathered}$$

In the x-axis, there is no acceleration :

$$\begin{gathered} v_{f_x}²=v_x²+2a_x∆x \\ \Rightarrow v_{f_x}²=v_x² \\ \Rightarrow v_{f_x}=v_x \\ =\mathbf{}\mathbf{6}\mathbf{.}\mathbf{06}\frac{\mathbf{m}}{\mathbf{s}} \end{gathered}$$

Calculating how far from the end of the ramp does the skateboarder touch down  

Calculate the distance between the skater and the ramp. The $∆y=-1.0m$is known.

$$\begin{gathered} ∆y=v_{f_y}t-\frac{1}{2}g{t}^2 \\ \Rightarrow -1=2.71t-\frac{1}{2}(9.80)\left(t^2\right) \\ \Rightarrow 0=4.9t^2-1.36t-1 \end{gathered}$$

To find$t$, use the quadratic equation. Ignore the negative value and use the positive value of$t=0.61s$ instead. :

role="math" localid="1650777944774" $$\begin{gathered} ∆x=v_{f_x}t+\frac{1}{2}ax{t}^2 \\ =(6.06\frac{m}{s})(0.61s)+\frac{1}{2}\left(0\frac{m}{s^2}\right){(0.61s)}^2 \\ =\boxed{\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\mathit{m}} \end{gathered}$$