Asked by Becca

An artillery shell is fired at an angle of 67.4◦ above the horizontal ground with an initial speed of 1560 m/s.
The acceleration of gravity is 9.8 m/s2 .
Find the total time of flight of the shell, neglecting air resistance. Find its horizontal range, neglecting air resistance
Answered by Henry
Vo = 1560 m/s @ 67.4o

Range = Vo^2*sin(2A)/g
= 1560^2*sin(134.8)/9.8 = 176,205 m.

Range = Vo*cos67.4*T = 176,205
T = 176,205/1560*cos67.4 = 294 s. in
flight.

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