Substituting g= 10 and v0= 10 into s=v0t- 12gt2, we get:
s =-5t 2+ 10t =-5(t- 1)2+5，
It is a parabola with a downward opening,
So the maximum value is 5, and 5+2 = 7m above the ground.
2. As shown in the figure, in the plane rectangular coordinate system, the image of the quadratic function y=x2+bx+c intersects with the X axis at points A and B, with point A on the left side of the origin, the coordinate of point B being (3,0), intersecting with the Y axis at point C (0,3), and point P being the moving point on the parabola below the straight line BC.
(1) Find the expression of this quadratic function.
(2) Connect PO and PC, and fold △POC along Co to get quadrilateral POP'C. Is there a point P that makes quadrilateral POP' C a diamond? If yes, request the coordinates of point P at this time; If it does not exist, please explain why.
(3) When the point P moves to what position, the area of the quadrilateral ABPC is the largest, and the coordinates of the point P and the maximum area of the quadrilateral ABPC are obtained.
1, the analytical formula of quadratic function can be obtained from the problem: y=x? -2x-3；
2. Set point P coordinate (x, y),
When the quadrangle is a diamond,
po = pc，PP' ⊥OC,OC=3，
∴ yP=-3/2，
When yP=-3/2,
-3/2=x？ -2x-3，
Found: x = √ 10/2+ 1,
∵x & gt； 0,
∴x= 1+√ 10/2，
∴ Point P coordinate (1+√ 10/2,-3/2);
3. Let the area be s, and the area of quadrilateral ABPC =S△ABC+S△BPC.
The vertical line of the x-axis intersection p intersects BC at point q,
Then PQ=X-3-(x? -2x-3)
=-x？ +3 times
s△BPC = 1/2 *(x？ +3x)*3
=-3/2(x？ -3 times)
∴S=6-3/2(x？ -3 times)
=-3/2(x-3/2)？ +6+27/8
=-3/2(X-3/2)？ +75/8
When x=3/2,
The maximum area of s is 75/8, and the p coordinate is (3/2,-15/4).
3. In the right triangle ABC, ∠ c = 90, AC = 4°, BC = 3°, and a plane rectangular coordinate system is established with A as the origin and the straight line of AC as the X axis, then the coordinate of point B is (-3,4).
: ∫∠C = 90, AC=4, BC=3, with A as the origin,
The coordinate of point ∴c is (-3,0),
BC = 4，
∴ The coordinates of point B are: (-3,4).
So the answer is: (-3,4).
4. As shown in the figure, it is a cylindrical water cup with a lid, with a bottom circumference of 6πcm and a height of 18cm. If the overlapping part between the lid and the cup body is ignored, the material needed to make a cup like 10 is at least ().
a、 108πcm2 B、 1080πcm2 C、 126πcm2 D、 1260πcm2
Solution: Let the base radius be r,
Then 2πr=6π,
The solution is r=3,
∴ The bottom area is 9π,
The transverse area is 6π× 18= 108π.
The surface area of a cup is 108π+2×9π= 126π,
∴ To make a water cup like 10, the minimum required material is10×126π =1260π.
Choose D.