Question

A motorcyclist attempts to leap a 48-m-wide river gorge. The point where he leaves, the ground is sloped upward at an angle of 15o to the horizontal. The landing point is 5.9 m lower than the takeoff point. (a) What is the minimum speed necessary for the cyclist to clear the jump? (b) If the cyclist exceeds that minimum speed by 50%, how far from the far edge will he land? {25 m/s, 44 m}

Answer 1

PROJECTILE

1)

along horizontal
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initial velocity v0x = v*cos(theta)

acceleration ax = 0

initial position = xo = 0

final position = x = 48

displacement = x - xo

from equation of motion

x - x0 = v0x*T+ 0.5*ax*T^2

x - x0 = v*costheta*T

T = (x - x0)/(v*cos(theta))......(1)

along vertical
______________

v0y = v*sin(theta)

acceleration ay = -g = -9.8 m/s^2

initial position y0 = 0 m

final position y = -5.9

from equation of motion

y-y0 = vy*T + 0.5*ay*T^2 .........(2)

using 1 in 2

y-y0 = (v*sin(theta)*(x-x0))/(v*cos(theta)) - (0.5*g*(x-x0)^2)/(v^2*(cos(theta))^2)

y-y0 = (x-x0)*tan(theta) - ((0.5*g*(x-x0)^2)/(v^2*(cos(theta))^2))

-5.9 -0 = (48*tan15) - (0.5*9.8*48^2/(v*cos15)^2)

v = 25.4 m/s

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for vnew = 1.5*v = 1.5*25.3 = 37.5 = 38 m/s

-5.9 = (x*tan15) - (0.5*9.8*x^2/(38*cos15)^2)

x = 91.6 m

He will land at 91.4 - 48 = 44 m