Let's write the equations of motion of the football in horizontal and vertical directions:
x=v0tcosθ,(1)y=v0tsinθ−21gt2,(2)here, x is the horizontal displacement of the football, v0=23 sm is the initial velocity of the football, t is the total time of flight of the football, θ=20∘ is the launch angle, y is the vertical displacement of the football (or the height) and g=9.8 s2m is the acceleration due to gravity.
a) Let's find the horizontal and vertical components of the football's initial velocity:
v0x=v0cosθ=23 sm⋅cos20∘=21.6 sm,v0y=v0sinθ=23 sm⋅sin20∘=7.87 sm.b) Let's first find the time that the football takes to reach the highest peak of the trajectory:
vy=v0sinθ−gtrise,0=v0sinθ−gtrise,trise=gv0sinθ=9.8 s2m23 sm⋅sin20∘=0.8 s.
c) Then, we can find the total time of flight of the footbal:
t=2trise=2⋅0.8 s=1.6 s.d) Finally, we can substitute trise into the second equation and find the maximum height:
ymax=v0sinθ⋅gv0sinθ−21g(gv0sinθ)2,ymax=2gv02sin2θ,ymax=2⋅9.8 s2m(23 sm)2⋅sin220∘=3.16 m.Answer:
a) v0x=21.6 sm,v0y=7.87 sm.
b) trise=0.8 s.
c) t=1.6 s.
d) ymax=3.16 m.
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