Drag Race Problem

A Drag Racer accelerates at a uniform rate from it's starting point.
It travels the last 1/4 of the distance from the starting point
to the finish line in 3 seconds.
How long did it take to travel the entire
distance from starting point to finish line?

This uses the Quadradic Formula:
A*X*X + B*X + C = 0
=> X = (-B + Sq-Rt(B*B - 4*A*C))/(2*A)
=> X = (-B - Sq-Rt(B*B - 4*A*C))/(2*A)

Enter: Enter .25 = 1/4 of distance here = y1
Enter: Enter 2.5 seconds here = y2
Enter: Enter 2640 = 1/2 mile ** (1320 = 1/4 mile) ** (2640 = 1/2 mile) ** (3960 = 3/4 mile) = y3
Enter: Enter 3200 lbs = weight of the car
Enter: Enter .07 = Friction of Car on the Road

Answer: b1 = 1 - y1 = .75
Answer: b2 = 1/b1 = 1.333333
Answer: b3 = b2 - 1 = .3333333
Answer: b4 = (2*y2)*(b2)*(-1) = -6.66666
Answer: b5 = y2*y2*b2 = 8.33333
Answer: b6 = b4*b4 - 4*b3*b5 = 33.33333
Answer: b7 = Sq-Rt(b6) = 5.77350269
Answer: b8 = (-b4 + b7)/(2*b3) = Time to travel the whole distance = 18.660254 seconds

Answer: b9 = (-b4 - b7)/(2*b3) = 1.33974596
Answer: b10 = y3/b8 = V2 = Ft/Sec = Final Speed = 141.47717 ft/sec
Answer: b11 = b10/b8 = a = V2/T = 7.581738 ft/sec/sec

Answer: b12 = 88/b11 = Time to Reach 60 mph = Seconds = 11.60683 seconds
Answer: b13 = b10*(60/88) = V2 = Final Speed = mph = 96.4617 mph

Answer: b14 = y4*y5 = 3200*.07 = 224 lbs to roll car
Answer: b15 = y4/32 = 100 slugs = mass of the car
Answer: b16 = b15*b11 = ma = Force = 758.173888 lbs
Answer: b17 = b14 + b16 = 758.17 + 224 = 982.1738 = Total Force pushing the car
Answer: b18 = (88*88)/(2*b11) = (V2*V2)/(2*a) = X = Distance traveled to reach 60 mph = 510.7007 ft
Answer: b19 = b17*b18 = Ft-Lbs = Force times Distance = F*X = Work Done = 501596.9758 ft-lbs
Answer: b20 = b19/b12 = (Ft-Lbs)/Second = Power = 43215.65108 (Ft-Lbs)/Second
Answer: b21 = b20/550 = HP = Horse-Power (550 Ft-Lbs per second = 1 HP) = 78.5739 HP Developed

A helium baloon carrying a baseball goes up
to a height h feet where it pops and
releases the ball. The ball travels the
last 3000 feet in 5 seconds.
Assume no friction and acceleration
due to gravity to be 32 ft/sec/sec.
How high was the ball when it dropped?

Horizon Problem

Fill in the information and then click on button.

Enter: Eye Level Height = 5280 feet
Enter: Radius of the Earth = 3959 miles

Answer: Distance to Horizon = Miles = 88.98876333560322 miles
Answer: Distance to Horizon = Feet = 469860.670411985 feet
Formula used here:
Let d*d = 2*R*h + h*h
=> d = SqRt(2*R*h + h*h)
R = 3959 miles
h = 30/5280 miles
This is distance to tangent of circle!
When d = 89 miles you will notice
that h = 1 mile!
So every time you travel 89 miles
you should drop 1 mile because of the
curve of the earth!
=> C = 2*Pi*R = 24875 miles
=> 24875/89 = 280 pie slices
=> if you traveled
1/4 of the circle = 6218.75 miles
=> 280/4 = 70 pie slices (each a mile drop)
=> 70 mile drop which cannot be right.
Since 1/4 of the circle would result
in a drop of the Radius = 3959 miles?

Rule of thumb approximation:
Let h = eye level height = feet
Nautical miles d = 1.17*(SqRt(h))
1 N.Mile = 6076 feet and 1 S.Mile = 5280 feet