Vo = 15.4m/s[38.7o]
Xo = 15.4*cos38.7 = 12 m/s.
Yo = 15.4*sin38.7 = 9.63 m/s.
Range = Vo^2*sin(2A)/g
Range = 15.4^2*sin(77.4)/9.8 = 23.62 m.
Range = Xo * T = 23.62 m.
Range = 12 * T = 23.62
T = 1.97 s. in air.
d=Vr * T = 23.62-17.7 = 5.92 m. to run.
Vr * 1.97 = 5.92
Vr = 3.0 m/s. = Speed of the receiver.
A football is thrown toward a receiver with
an initial speed of 15.4 m/s at an angle of
38.7
◦
above the horizontal. At that instant,
the receiver is 17.7 m from the quarterback.
The acceleration of gravity is 9.81 m/s
2
.
With what constant speed should the receiver run to catch the football at the level at
which it was thrown?
Answer in units of m/s
1 answer
0 answers