Sample+Answer+Letter


 * Sample Answer Letter:**

Mr. I. C. Light, Engineer SpectroMagic Fireworks Corp. 123 Benjamin Franklin Parkway Philadelphia, PA 19102

April 19th, 2011

Ms. Ginny Sparks, Vice President of Operations SpectroMagic Fireworks Corp. 123 Benjamin Franklin Parkway Philadelphia, PA 19102

Dear Ms. Sparks:

I am addressing your questions with regard to the SpectroMagic Fireworks Show we are planning for July 4th, 2010. The timing of the explosions is more important than their height, as we want to coordinate the explosions with music. However, we still need to know the height of the explosion, as the firework explodes at approximately its highest point. When the highest point is found, the value of t (time) at this point is the time at which the firework will explode.

Through the following calculations, I have discovered that a firework will explode 4 seconds after it is launched and that the height of the firework when it explodes is 80 meters. Therefore, we will need to launch the individual fireworks 4 seconds prior to starting the music and 4 seconds prior to any select points in the music you wish to highlight with firework explosions.

My calculations are as follows: 1) The equation I derived for the time-height relationship of the firework is h=(-4.9t2+39.2t+1.6). This is in the format of parabolic equations: y=ax2+bx+c, where a ≠0 and x (axis of symmetry) = -b/2a. 2) Given that we know that the highest point of the explosion would occur at the vertex of the parabola made by projecting the firework into the sky, we would solve for the t coordinate of the vertex, which is the time. This t coordinate is on the axis of symmetry of this parabola. Since t=-b/2a or –39.2/(2*(-4.9) = -39.2/-9.8 =4. Thus, the firework will explode when it reaches the vertex or when t=4 seconds. 3) Furthermore, the height of the firework when it explodes is the height when t=4. Thus, h=-4.9(42) + 39.2 (4) +1.6, or 80 meters.

You also requested information on the debris path of our fireworks display. The path of the firework debris can be modeled by the parabolic equation height = -.04x2 + 2x + 8, (where x is the horizontal distance in feet). Thus, we need to solve for the horizontal distance from the launch when the debris hits the ground or when h=0. Hence, we are looking to solve for x (horizontal distance in feet) in the equation 0=-.04x2 + 2x + 8. I have calculated that the debris will land about 53.7 feet from the launch site. My work is as follows:
 * < -0.04x2 + 2x + 8=0 ||< equation for where debris will land ||
 * < __-0.04x2 + 2x + 8=0__ ||< divide each side by -0.04 ||
 * < -0.04 -0.04 ||<  ||
 * < x2 – 50x – 200 = 0 ||< simplify ||
 * < x2 – 50x – 200 + 200 = 0 + 200 ||< add 200 to each side ||
 * < x2 – 50x = 200 ||< simplify ||
 * < x2 – 50x + 625 = 200 + 625 ||< Since (50/2) 2 = 625, add 625 to each side (completing the square) ||
 * < x2 – 50x + 625 = 825 ||< simplify ||
 * < (x – 25) 2 = 825 ||< factor x2 –50x + 625 ||
 * < x – 25 = + or – the square root of 825 ||< take the square root of each side ||
 * < x – 25 + 25 = + or – (square root of 825) + 25 ||< add 25 to each side ||
 * < x = 25 + or – (square root of 825) ||<  ||
 * < Thus, x = 53.7 or –3.7 feet. ||< Since we are looking for a distance, we can ignore the negative number. Thus the debris will land about 53.7 feet from the launch site. ||

My calculations indicate that spectators must stand at least 53.7 feet away from the launch site. That is where barricades should be put up at Penn’s Landing. Attached please find my design for the poster depicting the fireworks projected path and the distance people must stand away from the launch site. Any questions, feel free to contact me. I will work with the music planners to coincide the fireworks with the music you end up choosing.

Sincerely,

Mr. I. C. Light, Engineer