Arookie quarterback throws a football with an initial upward velocity component of 12.0 m> s and a horizontal velocity component of 20.0 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? how does this compare with the time calculated in part (a)? (d) how far has the football traveled horizontally during this time? (e) draw x-t, y-t, [email  protected], and [email  protected] graphs for the motion.

Draw a diagram to illustrate the problem as shown below.

The ball is thrown from A, rises to a maximum height at B, and returns to the original level at C.
Given:
v = 12.0 m/s, vertical component of launch velocity
u = 20 m/s, horizontal component of launch velocity.
Note that g = 9.8 m/s², the acceleration due to gravity.
Wind resistance is ignored.

Part a.
The time, t, required to reach B from A is given by
0 = 12 - 9.8t
t = 12/9.8 = 1.2245 s

The time is 1.225 s (nearest thousandth)

Part b.
The height attained is
h = 12*1.2245 - 05*9.8*1.2245²
   = 7.347 m

The height attained is 7.35 m (nearest hundredth) above A.

Part c.
When the ball returns the level of point A or C at time t,
0 = 12t - 0.5*9.8t²
t(12 - 4.9t) = 0
t = 0 (point A), or
t = 4.9/2 = 2.45 s (point C)
Notice that 2.45/2 = 1.225 s, which is equal to the time to reach B from A.

The time taken to travel from A to C is  2.45 s.
This time is twice the time taken to travel from A to B.

Part d.
The horizontal distance traveled from A to C is
d = 20*2.45 = 49 m

49 m

Part e.
x(t) = 20t  for horizontal travel
y(t) = 12t - 4.9t²  for vertical travel
vy(t) = 12 - 9.8t  for vertical velocity
vx(t) = 20  for horizontal velocity.
Graphs of x-t, y-t, vx-t, vy-t are shown below.

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