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

Answer:

60.07 m

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