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.

The position P is:

ft     Remember that the position is a vector. Observe the attached image

Step-by-step 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 is:

Where 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:

ft/sec

ft/sec

Thus

The height after 2 sec is:

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

Where

for this case

ft / sec

ft/sec

So

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

Finally the vector position P is:

ft

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