# physics

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a football is thrown to a moving receiver .the football leaves the quarterback's hands 1.75m above the ground with a velocity of 17.0m/s[25deg]if the receiver starts 12.0m away from the quarterback along the line of flight of the ball when is is thrown , what constant velocity must she have to get to the ball at the instant it is 1.75 m above the ground ?

• physics -

Vo = 17m/s[25o]
Xo = 17*cos25 = 15.4 m/s.
Yo = 17*sin25 = 7.2 m/s.

Y = Yo + g*t = 0 @ max. Ht.
Tr = -Yo/ g = -7.2/-9.8 = 0.73 s. = Rise
time.

Tf = Tr = 0.73 s. = Fall time.

D = Xo*(Tr+Tf) = 15.4 * 1.46 = 22.5 m.

V=(D-12)/(Tr+Tf)=(22.5-12)/1.46=7.19m/s

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