Arookie quarterback throws a football with an initial upward velocity component of 16.1 m/s and a horizontal velocity component of 18.2 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)t = 1.61 sec.

b)h = 12. 9 m

c)The time required to return on the initial position (T)= 3.22 sec

d)This time is 2 times of the time calculated in part a.

e)The horizontal distance = 58. 64 m

Explanation:

Given that

Upward velocity component(Vo) = 16.1 m/s

Horizontal velocity component(Uo) = 18.2 m/s

a)

When the upward component will become zero then this will be on the highest point

We know that

V= Vo - g.t

0= 16.1 - 10 x t

t = 1.61 sec.

b)

h = 12. 9 m

c)

The time required to return on the initial position = 2 x t

The time required to return on the initial position(T) = 2 x 1.61 = 3.22 sec

The time required to return on the initial position (T)= 3.22 sec

d)

This time is 2 times of the time calculated in part a.

e)

Range R = Uo .T

R = 18.2 x 3.22

R = 58.64 m

The horizontal distance = 58. 64 m

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