Arookie quarterback throws a football with an initial upward velocity component of 17.0 m/s and a horizontal velocity component of 18.3 m/s . ignore air resistance. a. how much time is required for the football to reach the highest point of the trajectory?
b. how high is this point?
c. how much time (after it is thrown) is required for the football to return to its original level?
d. how does this compare with the time calculated in part (a).
e. how far has it traveled horizontally during this time?

(a) 1.73 s

(b) 14.75 m

(c) 3.36 s

(d) double

(e) 63.32 m

Explanation:

Vertical component of initial velocity, uy = 17 m/s

Horizontal component of initial velocity, ux = 18.3 m/s

(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.

Use first equation of motion in vertical direction

vy = uy - gt

0 = 17 - 9.8 t

t = 1.73 seconds

(B) Let the highest point is at height h.

Use III equation of motion in vertical direction

0 = 17 x 17 - 2 x 9.8 x h

h = 14.75 m

(C) The time taken by the ball to return to original level is T.

Use second equation of motion i vertical direction.

h = 0 , u = 17 m/s

0 = 17 t - 0.5 x 9.8 t^2

t = 3.46 second

(D) It is the double of time calculated in part A

(E) Horizontal distance = horizontal velocity x total time

d = 18.3 x 3.46 = 63.32 m

what is the actual question though?