Asked by BN
If a rocket is launched at an initial speed of 120m/s at an angle of 25 degrees, find the time of flight, range, and it's highest height.
Answers
Answered by
Henry
Vo = 120m/s[25o].
Xo = 120*Cos25 = 108.8 m/s.
Yo = 120*sin25 = 50.71 m/s.
Yf = Yo + g*Tr = 0 at max h.
Tr = -Yo/g = 50.71/-9.8 = 5.17 s. = Rise time.
Tf = Tr = 5.17 s. = Fall time.
a. T = Tr + Tf = Time of Flight.
b. Range = Xo*T.
c. h max = Yo*Tr + 0.5g*Tr^2.
g = -9.8 m/s^2.
Xo = 120*Cos25 = 108.8 m/s.
Yo = 120*sin25 = 50.71 m/s.
Yf = Yo + g*Tr = 0 at max h.
Tr = -Yo/g = 50.71/-9.8 = 5.17 s. = Rise time.
Tf = Tr = 5.17 s. = Fall time.
a. T = Tr + Tf = Time of Flight.
b. Range = Xo*T.
c. h max = Yo*Tr + 0.5g*Tr^2.
g = -9.8 m/s^2.
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