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

He becomes a croco-dile food

Step-by-step 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 = (h)/(l)`

making
`l` the subject

`l = (h)/(sin \theta )`

substituting values

`l = (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= √(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 + (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

`(-b \pm √(b^2 -4ac) )/(2a)`

substituting values

`(-3.05 \pm √((3.05)^2 - 4(-4.9) * 2) )/(2 *( -4.9))`

`t = (-3.05 + 6.9)/(-9.8) \ or t = (-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