Respuesta :
Answer:
He becomes a croco-dile food
Explanation:
From the question we are told that
The height is h = 2.0 m
The angle is [tex]\theta = 20^o[/tex]
The distance is [tex]w = 10m[/tex]
The speed is [tex]u = 11 m/s[/tex]
The coefficient of static friction is [tex]\mu = 0.02[/tex]
At equilibrium the forces acting on the motorcycle are mathematically represented as
[tex]ma = mgsin \theta + F_f[/tex]
where [tex]F_f[/tex] is the frictional force mathematically represented as
[tex]F_f =\mu F_x =\mu mgcos \theta[/tex]
where [tex]F_x[/tex] is the horizontal component of the force
substituting into the equation
[tex]ma = mgsin \theta + \mu mg cos \theta[/tex]
[tex]ma =mg (sin \theta + \mu cos \theta )[/tex]
making a the subject of the formula
[tex]a = g(sin \theta = \mu cos \theta )[/tex]
substituting values
[tex]a = 9.8 (sin(20) + (0.02 ) cos (20 ))[/tex]
[tex]= 3.54 m/s^2[/tex]
Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as
[tex]sin \theta = \frac{h}{l}[/tex]
making [tex]l[/tex] the subject
[tex]l = \frac{h}{sin \theta }[/tex]
substituting values
[tex]l = \frac{2}{sin (20)}[/tex]
[tex]l = 5.85m[/tex]
Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as
[tex]v^2 =u^2 + 2 (-a)l[/tex]
The negative a means it is moving against gravity
substituting values
[tex]v^2 = (11)^2 - 2(3.54) (5.85)[/tex]
[tex]v= \sqrt{79.582}[/tex]
[tex]= 8.92m/s[/tex]
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
[tex]v_x = vcos 20^o[/tex]
substituting values
[tex]v_x = 8.92 * cos (20)[/tex]
[tex]= 8.38 m/s[/tex]
Initial velocity along the y-axis which is mathematically evaluated as
[tex]v_y = vsin\theta[/tex]
substituting values
[tex]v_y = 8.90 sin (20)[/tex]
[tex]= 3.05 m/s[/tex]
Now the motion through the pool in the vertical direction can mathematically modeled as
[tex]y = y_o + u_yt + \frac{1}{2} a_y t^2[/tex]
where [tex]y_o[/tex] is the initial height,
[tex]u_y[/tex] is the initial velocity in the y-axis
[tex]a_y[/tex] is the initial acceleration in the y axis with a constant value of ([tex]g = 9.8 m/s^2[/tex])
at the y= 0 which is when the height above ground is zero
Substituting values
[tex]0 = 2 + (3.05)t - 0.5 (9.8)t^2[/tex]
The negative sign is because the acceleration is moving against the motion
[tex]-(4.9)t^2 + (2.79)t + 2m = 0[/tex]
Solving using quadratic formula
[tex]\frac{-b \pm \sqrt{b^2 -4ac} }{2a}[/tex]
substituting values
[tex]\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}[/tex]
[tex]t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}[/tex]
[tex]t = -0.39s \ or \ t = 1.02s[/tex]
since in this case time cannot be negative
[tex]t = 1.02s[/tex]
At this time the position the motorcycle along the x-axis is mathematically evaluated as
[tex]x = u_x t[/tex]
x [tex]=8.38 *1.02[/tex]
[tex]x =8.54m[/tex]
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
