Afootball is thrown toward a receiver with an initial speed of 16.9 m/sat an angle of 36.5◦ above the horizontal. at that instant, the receiver is 16.8 m from the quarterback. the acceleration of gravity is 9.81 m/s2.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 = 5.39 m/s

Explanation:

vi = 16.9 m/s

∅ = 36.5º

d = 16.8 m

g = 9.81 m/s²

vr = ?

We have to get Xmax = R  as follows

R = vi²*Sin (2∅) / g

⇒  R = (16.9 m/s)²*Sin (2*36.5º) / (9.81 m/s²)

⇒  R = 27.842 m

Now we can get t, using the formula:

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

⇒   t = (27.842 m) / (16.9 m/s*Cos 36.5º) = 2.049 s

We get x (the distance which the receiver must be run) as follows

x = R - d

⇒  x = 27.842 m - 16.8 m = 11.042 m

Finally we have

vr = x / t

⇒  vr = 11.042 m / 2.049 s = 5.39 m/s

see earlier

80cm/s.

24cm/s.

53.3cm/s.