An object is launched upward at an initial velocity of 80 fe

An object is launched upward at an initial velocity of 80 feet per second from level. At what times is the object 64 feet above the ground? The area of a circle is 24 square inches. Find the radius of the circle.

Solution

3) h = ut - gt^2/2 where u = intial vel. ; h = height

h = 80t - 32.2t^2/2

h = 80t - 16.1t^2

h(t) = 64 ; solve for t :

64 = 80t - 16.1t^2

-16.1t^2 +80t -64 =0

solve for t : t = 1 sec ,3.96 sec height is 64 feet above ground

4) Area of circle = pi*(radius)^2

24 = pi(r)^2

r = sqrt( 24/pi) = 2.76 inches