A daredevil on a motorcycle leaves the speed of a ramp with a speed of 33.5 m/s

Question

A daredevil on a motorcycle leaves the speed of a ramp with a speed of 33.5 m/s as in the figure below. If his speed is 31.8 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance, m A 44.0-kg projectile is fired at an angle of 30.0 degree above the horizontal with an initial speed of 1 40 times 102 m/s from the top of a cliff 114 m above level ground, where the ground is taken to be y = 0. (a) What is the initial total mechanical energy of the projectile? J (b) Suppose the projectile is traveling 99.7 m/s at Its maximum height of y = 318 m. How much work has been done on the projectile by air friction? J (c) What is the speed of the projectile immediately before it hits the ground if air friction docs one and a half times as much work on the projectile when it is going down as it did when it was going up?

Explanation / Answer

4) form the conservation of energy

k1+u1 = k2+u2

here u1 = 0

mgh = 1/2m(v1^2-v2^2)

h = 1/2*9.8(33.5^2-31.8^2)

h = 5.66 m

5) Ei = k1+u1 = 1/2*44*140^2+44*9.8*114 = 431200+49156.8 = 480356.8 J

b) wf = Ef-Ei

Ef = 1/2*44*99.2^2+44*9.8*318 = 216494.08+137121.6 = 353615.68 J

wf = -126741.12 J

c) w tot = w+1.5 w = 2.5 wf = -3.17*10^5 J

w tot = k2+u2-k1-k1

w tot = 1/2m(v2^2-v1^2)-mgh1

v2 = sqrt(2wtot+mv1^2+2mgh1/m) = sqrt(-6.34*10^5+8.624*10^5+9.83*10^5/44)

v2 = 165.93 m/s

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