Question
A golf ball struck with an initial speed of 41.8m/s. What is the range of the shot if the launch angle is 30.0º?
Answers
Henry
Vo = 41.8m/s[30o]
Xo = 41.8*cos30 = 36.2 m/s.
Yo = 41.8*sin30 = 20.9 m/s.
Y = Yo + g*Tr = 0 @ max. ht.
Tr = -Yo/g == -20.9/-9.8 = 2.13 s. =
Rise time.
Tf = Tr = 2.13 s. = Fall time.
Range = Xo*(Tr+Tf) = 36.2m/s * 4.26s =
154.4 m.
Xo = 41.8*cos30 = 36.2 m/s.
Yo = 41.8*sin30 = 20.9 m/s.
Y = Yo + g*Tr = 0 @ max. ht.
Tr = -Yo/g == -20.9/-9.8 = 2.13 s. =
Rise time.
Tf = Tr = 2.13 s. = Fall time.
Range = Xo*(Tr+Tf) = 36.2m/s * 4.26s =
154.4 m.