Question

A daredevil decides to jump a canyon. To do so, she drives a motorcycle up an incline sloped at an angle of 35 degrees. If she leaves the incline at a speed of 25m/s what is the maximum width of the canyon that she can cross? (she lands and takes off at the same elevation.)

Answer 1

60.07 m

Step-by-step explanation:

`v` = speed of launch = 25 m/s

`\theta` = angle of incline = 35 deg

Consider the motion along the vertical direction :

`v_(oy)` = initial velocity along vertical direction =
`v Cos\theta` =
`25 Sin35` = 14.34 m/s

`a_(y)` = acceleration along vertical direction = - 9.8 m/s²

`t` = time of travel = ?

`y` = vertical displacement = 0 m

Using the kinematics equation

`y = v_(oy)t + (0.5)a_(y)t^(2)`

`0 = (14.34)t + (0.5)(- 9.8)t^(2)`

`t` = 2.93 sec

Consider the motion along the horizontal direction:

`v_(ox)` = initial velocity along horizontal direction =
`v Cos\theta` =
`25 Cos35` = 20.5 m/s

`a_(x)` = acceleration along vertical direction = 0 m/s²

`t` = time of travel = 2.93 sec

`x` = horizontal displacement = ?

Using the kinematics equation

`x = v_(ox)t + (0.5)a_(x)t^(2)`

`x = (20.5)(2.93) + (0.5)(0)(2.93)^(2)`

`x` = 60.07 m

`w` = width of canyon

`w` =
`x`

`w` = 60.07 m