Asked by Anonymous
An artillery shell is fired at an angle of 56.7 degrees
above the horizontal ground with an initial
speed of 1960 m/s. The acceleration of gravity is 9.8 m/s^2.
Find the total time of flight of the shell and the horizontal range,
neglecting air resistance.
above the horizontal ground with an initial
speed of 1960 m/s. The acceleration of gravity is 9.8 m/s^2.
Find the total time of flight of the shell and the horizontal range,
neglecting air resistance.
Answers
Answered by
Henry
Vo = 1960m/s[56.7o].
Xo = 1960*Cos56.7 = 1076 m/s
Yo = 1960*sin56.7 = 1638 m/s.
Y = Yo + g*Tr = 0.
1638 + (-9.8)Tr = 0,
Tr = 167.2 s. = Rise time.
Tf = Tr = 167.2 s. = Fall time.
T = Tr + Tf = 167.2 + 167.2 = 334.4 s. = Time in flight.
Range = Xo * T = 1076 * 334.4 =
Xo = 1960*Cos56.7 = 1076 m/s
Yo = 1960*sin56.7 = 1638 m/s.
Y = Yo + g*Tr = 0.
1638 + (-9.8)Tr = 0,
Tr = 167.2 s. = Rise time.
Tf = Tr = 167.2 s. = Fall time.
T = Tr + Tf = 167.2 + 167.2 = 334.4 s. = Time in flight.
Range = Xo * T = 1076 * 334.4 =
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