Question 620717: please show how to solve this I am confused...
Altitude of a launched object. The altitude of an object, in meters, is given by the polynomial
h + vt − 4.9t2,
where h is the height, in meters, at which the launch occurs, v is the initial upward speed (or velocity), in meters per second, and t is the number of seconds for which the object is airborne
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Altitude of a launched object. The altitude of an object, in meters, is given by the polynomial
h + vt − 4.9t2,
where h is the height, in meters, at which the launch occurs, v is the initial upward speed (or velocity), in meters per second, and t is the number of seconds for which the object is airborne
**
rewrite equation:
-4.9t^2+vt+h
You should see this is an equation of a parabola that opens downwards. (has a maximum)
Its standard form: y=-A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex.
To change given equation to this form, you must complete the square, from which you can see the coordinates of the vertex.
Given equation gives the height(m) of the airborne object at any time t(sec)
h(t)=-4.9t^2+vt+h
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