A daredevil jumps a canyon 10.9 m wide. To do so, he drives a car up a 24° incline. (The daredevil lands on the other side of the canyon at the same elevation as takeoff.) (a) What minimum speed must he achieve to clear the canyon?

Respuesta :

Answer:

Minimum speed required =  12 m/s

Explanation:

Let the speed be y, and  upward direction be positive

Consider the vertical motion of car:-

    Initial velocity, u = ysin24

    Acceleration, a = -9.81 m/s²

    We need to find time when displacement is zero, ( s = 0m )

    Substituting in s = ut + 0.5at²

                      0 = ysin24 x t + 0.5 x -9.81 x t²

                      4.905t² - ysin24 x t = 0

                       t (4.905t - ysin24 ) = 0

                        t = 0 or [tex]t=\frac{ysin24}{4.905}=0.083y[/tex]

  Here time = 0 is not considered,

          So t = 0.083 y

Consider the vertical motion of car:-

    Initial velocity, u = ycos24

    Acceleration, a = 0 m/s²

    Time, t = 0.083y

    We need to find speed when displacement is 10.9 m, ( s = 10.9 m )

    Substituting in s = ut + 0.5at²      

                            10.9 = ycos24 x  0.083y + 0.5 x 0 x (0.083y)²

                            0.076y² = 10.9

                                      y = 11.99 = 12 m/s

   Minimum speed required =  12 m/s