Question

Suppose a football is kicked with an initial velocity of 82 ft/sec., at an angle of

58° to the ground. What is the position of the ball after two seconds?

(Remember: Acceleration due to gravity is 32 ft/s².) Round your answer

to the nearest whole foot. Show all work.

Answer 1

The position P is:

`P = 87\^x + 75\^y` ft Remember that the position is a vector. Observe the attached image

Explanation:

The equation that describes the height as a function of time of an object that moves in a parabolic trajectory with an initial velocity
`s_0` is:

`y(t) = y_0 + s_0t -16t ^ 2`

Where
`y_0` is the initial height = 0 for this case

We know that the initial velocity is:

82 ft/sec at an angle of 58 ° with respect to the ground.

So:

`s_0 = 82sin(58\°)` ft/sec

`s_0 = 69.54` ft/sec

Thus

`y(t) = 69.54t -16t ^ 2`

The height after 2 sec is:

`y(2) = 69.54 (2) -16 (2) ^ 2`

`y(2) = 75\ ft`

Then the equation that describes the horizontal position of the ball is

`X(t) = X_0 + s_0t`

Where

`X_ 0 = 0` for this case

`s_0 = 82cos(58\°)` ft / sec

`s_0 = 43.45` ft/sec

So

`X(t) = 43.45t`

After 2 seconds the horizontal distance reached by the ball is:

`X (2) = 43.45(2)\\\\X (2) = 87\ ft`

Finally the vector position P is:

`P = 87\^x + 75\^y` ft