5. during the eruption of mount st. helens in 1980. debris was ejected at a speed of over 440 feet per second

(300 miles per hour). the elevation in feet above sea level of a rock ejected at an angle of 75° above

horizontal is given by the function y(1) --1612 +425t+8200 where t is the time in seconds after the

iption. the rock's horizontal distance in feet from the point of ejection is given by x(i) = 1131. if the rock

landed at an elevation of 6000 feet, what is the horizontal distance from the point of ejection to where it!

landed rounded to the nearest hundred feet?

  3500 ft

Step-by-step explanation:

We assume you intend the equations to be ...

Then we can solve the second equation for t and use that in the first equation. We can solve the first equation for y=6000.

  x = 113t

  t = x/113

  y(t) = y(x/113) = -16(x/113)² +425(x/113) +8200 = 6000

  (16/12769)x² -(425/113)x -2200 = 0 . . . . . . subtract 6000, multiply by -1

Using the quadratic formula, the positive solution is ...

  x = ((425/113) +√((-425/113)² -4(16/12769)(-2200))/(2(16/12769))

The fraction in the denominator can be inverted and used as a multiplier. We also use a√b = √(a²b) to simplify the radical.

  x = (48025 +√4104275825)/32

  x ≈ 3502.8

Rock was thrown about 3500 feet horizontally.