Aquarterback throws a football to a teammate. the football is 6.5ft above the ground when it leaves the quarterback's hand. his teammate catches it 3.5s later, at a height above the ground of 5 ft.

projectile motion formula

h(t) = -16t2 + vt + h0

h0 = 6.5

v = ?

h = 5 when t = 3.5

The value of velocity, V for the given projectile motion formular is 54.14 m/s

Using the projectile motion formula :

To obtain the value of v ; we input the given values into the motion equation :

h(3.5) = - 16(3.5²) + 3.5v + 6.5

h(3.5) = - 16(12.25) + 3.5v + 6.5

h(3.5) = - 196 + 3.5v + 6.5

Collect like terms

-3.5v = 6.5 - 196

-3.5v = - 189.5

Divide both sides by -3.5 to isolate v

v = (189.5 / 3.5)

v = 54.14

Therefore, the value of velocity, V for the given parameter values is 54.14

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the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!