# During the eruption of Mount St. Helens in 1980, debris was ejected at a speed of 440 feet per second

#1

During the eruption of Mount St. Helens in 1980, debris was ejected at a speed of 440 feet per second. The height in feet of a rock ejected at angle of 75 degrees by the equation y(t)=-16t^2+425t+8200, where t is the time in seconds after the eruption. Maximum height? Time it hit the ground? Time it maintains 9000 ft.?

Assuming the coordinate system is at the ground, you can calculate the maximum height by differentiating with respect to t and then make an equality where the rate of change of y is equal to zero.

y’=-32t+425=0

t≈13.3s

Hence,

y(13.3)≈11.0E3ft

For the time it hits the ground, again minding the assumption made initially, just make an equality where y is equal to zero:

-16t^2+425t+8200=0

Using basic rules to solve quadratic functions (laaazy right now thats why just discrimant away dood):

t≈39.7s

Note that the other isn’t mentioned since negative time isn’t considered here.

As for the 9000ft:

-16t^2+425t-800=0

t≈2.04s

or

t≈24.5s

There must be a question where they ask how far the rock reaches right? Or else all the other information is useless.