Answer:
The position P is:
[tex]P = 87\^x + 75\^y[/tex] 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 [tex]s_0[/tex] is:
[tex]y(t) = y_0 + s_0t -16t ^ 2[/tex]
Where [tex]y_0[/tex] 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:
[tex]s_0 = 82sin(58\°)[/tex] ft/sec
[tex]s_0 = 69.54[/tex] ft/sec
Thus
[tex]y(t) = 69.54t -16t ^ 2[/tex]
The height after 2 sec is:
[tex]y(2) = 69.54 (2) -16 (2) ^ 2[/tex]
[tex]y(2) = 75\ ft[/tex]
Then the equation that describes the horizontal position of the ball is
[tex]X(t) = X_0 + s_0t[/tex]
Where
[tex]X_ 0 = 0[/tex] for this case
[tex]s_0 = 82cos(58\°)[/tex] ft / sec
[tex]s_0 = 43.45[/tex] ft/sec
So
[tex]X(t) = 43.45t[/tex]
After 2 seconds the horizontal distance reached by the ball is:
[tex]X (2) = 43.45(2)\\\\X (2) = 87\ ft[/tex]
Finally the vector position P is:
[tex]P = 87\^x + 75\^y[/tex] ft
