# law of sines

Printable View

• Jun 27th 2006, 09:09 AM
Dora
Help!
A ditch 2.5 m wide crosses a trail bike path. An upward incline of 15o 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.)
• Jun 27th 2006, 10:31 AM
Soroban
Hello, Dora!

Quote:

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 $A$ at a 15° angle.

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

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

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

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

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

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

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

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