Asked by Mollie
A footballl is thrown directly toward a receiver with initial speed of 18.0 m/s At an angle of 35.0 degrees above the horizontal. At that instant, The receiver is 18.0 m from the quarterback. In what direction and with what constant speed should the receiver run to catch the football at the level at which it was thrown?
Answers
Answered by
Henry
Range = Vo^2*sin(2A)/g.
Range = 18^2*sin(70)/9.8 = 31 m.
Range = Xo*T = 31 m.
Range = 18*cos35*T = 31
T = 2.1 s. = Time in air.
d = V*T = 31-18 = 13 m.
V * 2.1 = 13
V = 6.18 m/s away from quarterback.
d = 31 - 18 = 13 m.
Range = 18^2*sin(70)/9.8 = 31 m.
Range = Xo*T = 31 m.
Range = 18*cos35*T = 31
T = 2.1 s. = Time in air.
d = V*T = 31-18 = 13 m.
V * 2.1 = 13
V = 6.18 m/s away from quarterback.
d = 31 - 18 = 13 m.
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