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.
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?
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