Question

A young man tries to relive Evel Knievel's attempt to jump over Snake River Canyon on a motorcycle. The barrel has a width of L = 400. m, and both edges are at the same height. The height of the launch ramp at one rim of the canyon is h = 8.00 m above the rim, and the angle of the final part of the ramp is 45.0 ° from the horizontal. What is the minimum speed necessary for the free youth to make the jump? Neglect the resistance of the air.

Answer 1

v_o = speed of launch

= Angle of launch = 45

Consider the motion along the horizontal direction

X = horizontal displacement = width of the barrel = L = 400 m

v_ox = component of velocity along horizontal direction = v_o Cos = v_o Cos45

t = time taken

a_x = acceleration = 0 m/s²

Using the kinematics equation

X = v_ox t + (0.5) a_x t²

400 = (v_o Cos45) t + (0.5) (0) t²

t = 400/(v_o Cos45)                                                       Eq-1

Consider the motion along the vertical direction

Y = vertical displacement = - h = - 8 m

v_oy = component of velocity along vertical direction = v_o Sin = v_o Sin45

t = time taken

a_y = acceleration = - 9.8 m/s²

Using the kinematics equation

Y = v_oy t + (0.5) a_y t²

- 8 = (v_o Sin45) t + (0.5) (- 9.8) t²

Using Eq-1

- 8 = (v_o Sin45) (400/(v_o Cos45)) + (0.5) (- 9.8) (400/(v_o Cos45))²

400 = (400 tan45) + (0.5) (- 9.8) (400/(v_o Cos45))²

v_o = 62 m/s