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: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

report!

explanation:

physics means: "the branch of science concerned with the nature and properties of matter and energy. the subject matter of physics, distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms."