Question

A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and land at ground level on the other side. He has done this stunt many times and approaches it with confidence. Unfortunately, the motor-cycle engine dies just as he starts up the ramp. He is going 11 m/s at that instant, and the rolling friction of his rubber tires (coefficient 0.02) is not negligible. Does he survive, or does he become croco-dile food? Justify your answer by calculating the distance he travels through the air after leaving the end of the ramp

Answer 1

Answer:

He becomes a croco-dile food

Explanation:

From the question we are told that

    The height is h = 2.0 m

      The angle is  $\theta = 20^o$

     The distance is  $w = 10m$

       The speed is  $u = 11 m/s$

       The coefficient of static friction is  $\mu = 0.02$

At equilibrium the forces acting on the motorcycle are mathematically represented as

        $ma = mgsin \theta + F_f$

where  $F_f$ is the frictional force mathematically represented as

            $F_f =\mu F_x =\mu mgcos \theta$

where $F_x$ is the horizontal component of the force

substituting into the equation

            $ma = mgsin \theta + \mu mg cos \theta$

            $ma =mg (sin \theta + \mu cos \theta )$

               making  a the subject of the formula

      $a = g(sin \theta = \mu cos \theta )$

          substituting values

      $a = 9.8 (sin(20) + (0.02 ) cos (20 ))$

        $= 3.54 m/s^2$

Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as  

             $sin \theta = \frac{h}{l}$

making $l$ the subject

          $l = \frac{h}{sin \theta }$

substituting values

        $l = \frac{2}{sin (20)}$

           $l = 5.85m$

Apply Newton equation of motion we can mathematically evaluate the  final velocity at the end of the ramp  as

      $v^2 =u^2 + 2 (-a)l$

  The negative a means it is moving against gravity

      substituting values

      $v^2 = (11)^2 - 2(3.54) (5.85)$

           $v= \sqrt{79.582}$

              $= 8.92m/s$

The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is

    Initial velocity along the x-axis which is mathematically evaluated as

          $v_x = vcos 20^o$

       substituting values

         $v_x = 8.92 * cos (20)$

              $= 8.38 m/s$

Initial velocity along the y-axis which is mathematically evaluated as

             $v_y = vsin\theta$

      substituting values

             $v_y = 8.90 sin (20)$

                  $= 3.05 m/s$

Now the motion through the pool in the vertical direction can mathematically modeled as

        $y = y_o + u_yt + \frac{1}{2} a_y t^2$

where $y_o$ is the initial height,

         $u_y$ is the initial velocity in the y-axis

    $a_y$  is the  initial  acceleration in the y axis  with a constant value of ($g = 9.8 m/s^2$)

at the y= 0 which is when the height above ground is zero

      Substituting values

              $0 = 2 + (3.05)t - 0.5 (9.8)t^2$

The negative sign is because the acceleration is moving against the motion

                 $-(4.9)t^2 + (2.79)t + 2m = 0$

   Solving using quadratic formula

              $\frac{-b \pm \sqrt{b^2 -4ac} }{2a}$

substituting values

             $\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}$

                $t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}$

                $t = -0.39s \ or \ t = 1.02s$

since in this case time cannot be negative

             $t = 1.02s$

At this time the position the  motorcycle along the x-axis is mathematically evaluated as

              $x = u_x t$

               x  $=8.38 *1.02$

                   $x =8.54m$

So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of  the pool