α=45∘,v0=22m/s,l0=40m,h0=7m
Answers:
a)h=12.347m
b)l=49.387m
c)H=7.603m
d)The arrow will drop inside the castle walls
Explanation:
x=v0xt,v0x=v0cosα
y=v0yt−2gt2,v0y=v0sinα
vx=v0x,vy=v0y−gt
a) Let t1 be time point, when the arrow reaches the maximum height.
At this point vy=0,v0sinα−gt1=0,t1=gv0sinα
h=v0yt−2gt2=v0sinαgv0sinα−2g(gv0sinα)2=2gv02sin2α
h=2gv02sin2α=2∗9.8222∗21≈12.347m
b) Let t2 be time point, when the arrow reaches the maximum range.
l=v0xt2=v0cosα2t1=v0cosαg2v0sinα=gv02sin2α
l=gv02sin2α=9.8222∗1≈49.387m
c) Let t3 be time point, when the arrow reaches the wall.
l0=v0xt3
H=v0yt3−2gt32=v0sinαv0cosαl0−2v02cos2αgl02=l0tanα−2v02cos2αgl02=40∗1−2∗2221/29.8∗402=7.603m
d) H>h0⇒ the arrow will be under the wall
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