Question

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

Answer 1

Answer:

The position P is:

$P = 87\^x + 75\^y$ 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 $s_0$ 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