Word problem help please.

Question

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 height in feet of a rock ejected at an angel of 75 degrees is given by the equation y(t)=-16t^2+425+8200, where  t is the time in seconds after the eruption. The rock horizontal distance in feet from the point of ejection is give by x(t)=113t. Assuming the elevation of the surrounding country side is 0 feet. what is the horizontal distance form the point of ejection to where the rock would have landed? Round your answer to the nearest foot.

 

Okay so, please try your best to explain where the numbers come from, and explain what you do with those numbers and why. I really need to understand how to do these and i just cant seem to teach myself! :C

Answer 1

The equations are:

 

y(t) = -16t² + 440*sin(75^o)*t + 8200 = -16t² + 425t + 8200

x(t) = 440*cos(75^o)*t = 113t

 

Where:

• -16 ft/s² = the acceleration due to gravity
• 440*sin(75^o)*t = 425t = the initial upward velocity times time
• 8200 = the initial height (summit of the mountain)
• 440*cos(75^o)*t = 113t = the initial horizontal velocity times time

 

Since the elevation of the country around the mountain is stated to be 0 feet, the rock hits the ground when y(t) = 0:

 

0 = -16t² + 425t + 8200

 

Use the quadratic formula to find the values of t when y(t) = 0:

 

t = (-425/-32) ± (1/32)√(425²-4(-16)(8200))

t = 13.3 ± 26.2

t = -12.9 and 39.5 sec

 

Since we don't care about negative times, the rock will hit the ground in t = 39.5 seconds.  So how far will the rock travel horizontally in 39.5 seconds?

 

x(t=39.5) = 113*(39.5) = 4463.5 feet ≅ 4464 ft

 

 

Answer 2

This question is a bit more involved than the last one, but I'll do my best. The rock hits the ground when y(t) = 0, that is, its vertical height is zero.

 

In order to find the time at which this happens, solve the y(t) equation given for t where y(t)=0

(I'm guessing there is supposed to be a t multiplied by the second term since as written, it should simply say

y(t) = -16t² + 8625 where I just added 425+8200.)

 

0 = -16t² + 425t + 8200

 

Using the quadratic formula

 

If ax² + bx + c = 0, then x = (-b ± sqrt(b²-4ac))/(2a)

 

In your problem, a=-16, b=425, and c=8200, and x=t

 

Then, t= (-425 ± sqrt(425²-(4)(-16)(8200)))/(2*(-16))

 

Simplifying, we get t = (-425 ± 839.9)/(-32)

 

Since the trajectory of the debris is parabolic, there are 2 roots of the equation.

 

t= (-425 -839.9) / (-32) = 39.5 seconds

 

OR

 

t = (-425 +839.9) / (-32) = -13.0 seconds

 

The debris started moving when t=0 and therefore could not have hit the ground at a negative t.

 

Based on this, t=39.5 seconds.

 

Now, plug in t into the x(t) equation given

 

x(t) = 39.5 * 113 = 4464 feet