college algebra question

Question

A varsity soccer player kicks a soccer ball approximately straight upward with an initial velocity of 68 ft/sec. The ball leaves the players foot at a height of 3 feet. 

Part 1 of 2: (2 pts)

Use the position function to write a model representing the height of the soccer ball (in feet) t seconds after being kicked. (Assume the acceleration due to gravity is 32 ft/sec²). Be sure to reduce your first term.

Part 2 or 2: (6 pts)

Use that model to determine the times at which the ball is 73 feet in the air. Use FACTORING to solve. Be sure to write a sentence that explains both of your answers.

Answer 1

A)

Position function: s(t) = 1/2 at² + v_it + S_o

s(t) = -16t² + 68t + 3

B)

73 = -16t² + 68t + 3

(0 = -16t² + 68t - 70)/-16

0 = t² - 17/4 t - 35/8

0 = (t -7/4)(t - 5/2)

t = 7/4 seconds and t = 5/2 seconds

Answer 2

Many ways to skin a cat if you know some physics and the vertex form of the quadratic equation.

1) what goes up must come down and as such has an apogee in its flight path.

At apogee the velocity in the vertical direction goes to zero and things get very simple.

2) also one can plot y(t) in the vertex form generally as y(t) = a(t-h)² +k where at t=h the parabola vertex is the value of the height of our ball at the vertex....h is then the time elapsed for the ball to reach the vertex and k is the height of the ball.

velocity for constant acceleration at time t is v(t) = v₀ + at see we have with our knowledge of things attacked this without using the standard form of the quadratic.

At apogee v(t) is zero and t = - vo/a a is in the opposite direction of vo.

specifically vo = 68 ft/sec and a = -32 ft/sec-sec so t = 68/32 sec = 17/8 of a second and we know h in our parabolic vertex equation is 17/8, and a is -16 we also know y(17/8) = K.

Rather than substituting 17/8 for t in the standard quadratic an calculating K we can match the general forms to both arrive at h and K in the vertex form from the standard form:

standard form y = ax² +bc +c

vertex form y = a(x-h)² + K

then matching terms h = - b/2a which is a term one sees in solving for roots

and c = ah² + K or K = c- ah² = c - b²/4a

remember the discriminant b² -4ac.... -4aK = b² -4ac

so in our specific problem c = 3 ft the initial height of our ball

vo =b = 68 ft/sec

a = -32 ft/sec-sec/2 = -16 ft/sec-sec

b²/4a = (4x17)²/4(-16) = - (4x17)²/4²x4 = -17²/4 ft

k =( 17²/8 + 3) ft = 72.25 ft + 3 ft = 75 1/4 ft = 301/4 ft

We can write the vertex form y(t) = -16(t-17/8)² + 301/4 ft

But what does this tell us .....

1) after 17/8 of a second has elapsed the ball is at a height of 75.25 feet and at apogee.

The standard equation still pertains for displacement of our ball y= yo +vo t + 1/2 a t²

and here is the physics vo =0, yo =75.25 ft and 73 ft = 75.25 ft -16 t² where t is the time elapsed from apogee to when the ball is again at 73 ft from the ground.

2)

so t² = (9/4)/4x4 = (3/8)² and t = 3/8 sec...

so the total time elapsed from start of kick to apogee and then ball falling to 73 feet is 17/8 + 3/8 = 20/8 = 5/2 sec.

The parabola is symmetric and thus at 17/8 sec minus 3/8 of a second the ball was also 73 ft from the ground = 14/8 = 7/4 sec

What is not so obvious is that the ball passes through the plane in the horizontal at 73 feet 3/8 of a second before reaching the apogee and 3/8 of a second after reaching the apogee.

One can also then say the ball is at over 73 feet from the ground for 3/4 of a second.

The total time the ball is over 3 feet from the ground from the time of the kick is 2 x 17/8 or 17/4 seconds...

4 1/4 seconds.

The fraction of time the ball is over 73 feet is 3/4 sec/ 17/4 sec = 3/17 or 17 percent of its flight time.

one can even calculate the time required for the ball to hit the ground.

the distance to the ground is 3 ft set y(t) = 0 and v = -68 ft/sec

0 = 3- 68t -16t² or perhaps easier use our height at apogee 0 = 75.25 ft -16t²

17.34/8 of a second our 17/8 of a second plus 34/800 sec or 17/400 sec to drop that last 3 feet.

the total time of flight of our ball is 17/4 and 17/400 of a second.