Question
Calculate the hang time of a person who moves horizontally 2.00 m during a 1.23 m high jump.
Express your answer with the appropriate units
Express your answer with the appropriate units
Answers
Elena
The motion of jumping person is similar to projectile motion:
H=(Vo)^2•(sinα)^2/2g
L=(Vo)^2•(sin2α)/2g=
=(Vo)^2•(2sinα•cosα)/2g.
2H/L=sinα/cosα,
tanα=2H/L=21.23/2=1.23
α=arctan1.23=50.9°
Vo=sqr(2gH)/ sinα=6.33 m/s.
t=2Vo•sinα/g=1 s.
H=(Vo)^2•(sinα)^2/2g
L=(Vo)^2•(sin2α)/2g=
=(Vo)^2•(2sinα•cosα)/2g.
2H/L=sinα/cosα,
tanα=2H/L=21.23/2=1.23
α=arctan1.23=50.9°
Vo=sqr(2gH)/ sinα=6.33 m/s.
t=2Vo•sinα/g=1 s.