Afootball is thrown toward a receiver with an initial speed of 15.4 m/s at an angle of 36.2 ◦ above the horizontal. at that instant, the receiver is 15.8 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

vr = 3.906 m/s

Explanation:

vi = 15.4 m/s

∅ = 36.2º

d = 15.8 m

g = 9.81 m/s²

vr = ?

We have to get Xmax = R as follows

R = vi²*Sin (2∅) / g

⇒ R = (15.4 m/s)²*Sin (2*36.2º) / (9.81 m/s²)

⇒ R = 23.0437 m

then we can get t applying the equation

R = vi*Cos ∅*t    ⇒   t = R / (vi*Cos ∅)

⇒   t = (23.0437 m) / (15.4 m/s*Cos 36.2º) = 1.85 s

Now, we find the distance that the receiver must be run

x = R - d

⇒  x = 23.0437 m - 15.8 m = 7.2437 m

Finally, we get the speed

vr = x / t

⇒  vr = 7.2437 m / 1.85 s = 3.906 m/s

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