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- bob bundy
- 2013-03-01 20:30:43
75 is what I make it too. 
Bob
- 295Ja
- 2013-03-01 20:19:32
Thanks bob bundy! 
In Q1,using integration, I got the same answer as when I plugged in the values in the position function you gave but could you still
check my solution for any flaw?
Here it is:
v(t)=-10t+C*
s(t)=-5t²+C*t+C**
Given that speed=10 and since speed=absolute value of velocity, v=-10
v(0)=-10t+C*=-10
C*=-10
s(t)=-5t²-10t+C**
but s(3)=0
C**=75
s(t)=-5t²-10t+75
s(0)-s(3)=75
- bob bundy
- 2013-03-01 09:22:58
hi 295Ja
Q1. You can use
directly, where g = 10 t = 3 and u = 10
This assumes s(0) = 0
Q2 The picture below is my view of this problem.
On the left we have the graph of the function for the base.
On the right a 3D represenation of the base with one slice shown rising up in a third dimension.
The slice has area half y^2 and thickness dx
So integrate these to get the volume.
Bob
- 295Ja
- 2013-03-01 03:52:48
Hi! Help please.
Problem 1:
A ball was thrown downward from the top of a building at a speed of 10 meters per
second and it hit the ground 3 seconds later. How tall was the building? [Note: Use
-10 m/s² for acceleration due to gravity.]
>>Here's what I think I should do:
1. Find the velocity then position functions
Here's what I got:
Velocity Function:
v(t)=-10t+C*
Position Function:
s(t)=-5t²+C*t+C**
[Note: C*=constant after integrating a(t) and C**= constant after integrating v(t)]
2.Find the value of C* and C**
My question: is it right to use these: v(3)=0 and s(3)=0 to find them?
3.Evaluate s(0)-s(3)
Problem 2
The base of a solid S is the region bounded by y =square root of (16-x²) and the x-axis, and each
cross-section perpendicular to the x-axis is an isosceles right triangle having one leg on
the base of S. Using the method of slicing, determine the volume of S.
>>I have a hard time with this problem. A help on visualizing the figure and/or setting-up the integral that will give the volume would do.
Thank you!