Hello, Dora!

A ditch 2.5 m wide crosses a trail bike path.

An upward incline of 15° has been built up on the approach

so that the top of the incline is level with the top of the ditch.

What is the minimum speed a trail bike must be moving to clear the ditch?

(Add 1.4 m to the range for the back of the bike to clear the ditch safely.)

Code:

* *
* *
* 15° *
A * - - - - - - - - - - - - - * - - * B
: 2.5 : 1.4 :

The bike is launched at $\displaystyle A$ at a 15° angle.

It is to land at point $\displaystyle B\!,$ 3.9 meters to the right and at the same height.

The equations of projectile motion are: .$\displaystyle \begin{array}{cc}x\:=\v\cos15^o)t \\ y \:= \v\sin15^o)t - 16t^2\end{array}$

. . where $\displaystyle v$ is the initial velocity (launch speed).

We want $\displaystyle x = 3.9:\;\;(v\cos15^o)t\:=\:3.9$ **[1]**

and we want $\displaystyle y = 0:\;\;(v\sin15^o)t - 16t^2\:=\:0\quad\Rightarrow\quad (v\sin15^o)t\:=\:16t^2$ **[2]**

Divide **[2]** by **[1]**: .$\displaystyle \frac{(v\sin15^o)t}{(v\cos15^o)t}\:=\:\frac{16t^2} {3.9}\quad\Rightarrow\quad\tan15^o\:=\:\frac{16t^2 }{3.9}$

Solve for $\displaystyle t:\;\;t^2 \,= \,\frac{3.9\tan15^o}{16}\quad\Rightarrow\quad t \,= \,\frac{\sqrt{3.9\tan15^o}}{4} \,\approx \,0.25556 $ seconds.

From **[1]**, we have: .$\displaystyle v\:=\:\frac{3.9}{t\cos15^o}\:=\:\frac{3.9}{(0.2555 6333)\cos15^o} \:\approx\:15.8\text{ m/sec}$