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

answer: he got 23

explanation:

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

hope this ! this is not copied! search it up, it's free of plagiarism!