A football is thrown with an initial upward velocity component of 15.0m/s and a horizontal velocity component of 17.0m/s .

Question

a:How much time is required for the football to reach the highest point in its trajectory?
b:How high does it get above the ground?
c:How much time after it is thrown does it take to return to its original height?
d:How far has the football traveled horizontally from its original position?

Answer 1

Xo = 17 m/s.
Yo = 15 m/s.

a. Tr = -Yo/g = -15/-9.8 = 1.53 s. = Rise Time.

b. h = 0.5g*Tr^2 = 4.9*1.53*2 = 11.47 m.

c. T = Tr+Tf = 1.53 + 1.53 = 3.06 s.

d. Dx = Xo*T = 17m/s. * 3.06 = 52.02 m.

Answer 2

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