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Parabolic shot with friction coefficient?
1 Answer
- oubaasLv 71 day agoFavorite Answer
energy KE at slide top :
KE= Vo^2/2*m-m*g*L(sin 35+cos 45*μk)
KE = m(20^2/2-9.806*1.4*(0.574+0.819*0.18))= 190*m = m/2*V^2
V = √380 = 19.5 m/sec
h = 1.4*0.574 = 0.80 m
Δh = (V*sin 35)^2/2g = (380*0.574^2/19.6) = 6.39 m
H = h+Δh = 0.80+6.39 = 7.19 m
tdown = √2H/g = √2*7.19/9.806 = 1.21 sec
tup = V*sin 35/g = 19.5*0.574/9.806 = 1.14 sec
total evolution time t (from slide top on) = 1.21+1.14 = 2.35 sec
distance d = V*cos 35*t = 19.5*0.819*2.35 = 37.5 m
impact velocity Vi :
Vix = 19.5*0.819 = 16.0 m/sec
Viy = g*tdown = -9.06*1.21 = -11.97 m/sec
Vi = √Vix^2+Viy^2 = √16^2+11.97^2 = 19.9 m/sec ( a bit less than Vo)
heading angle = arctan Viy/Vix = arctan -11.97/16.0 = -36.8° ( 36.8° SoE)
