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

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#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.?

Answer:

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.