During a recent eruption, a piece of ash was ejected from a volcano with an initial velocity of 368 ft/sec. the height (h), in feet, of our ash projectile is given by the equation h=16t^2 + 368t where t is time, in seconds. the graph of the equation will be a parabola assume the volcano does not have a height. so, h= 0 and t=0 1.) when does the ash projectile reach it's maximum height? 2.)what is the maximum height? 3.)when does the ash projectile return to the ground?

For every corresponding pair of cross sections, the area of the cross section of a sphere with radius r is equal to the area of the cross section of a cylinder with radius and height 2r minus the volume of two cones, each with a radius and height of r. a cross section of the sphere is and a cross section of the cylinder minus the cones, taken parallel to the base of cylinder, is the volume of the cylinder with radius r and height 2r is and the volume of each cone with radius r and height r is 1/3 pie r^3. so the volume of the cylinder minus the two cones is therefore, the volume of the cylinder is 4/3pie r^3 by cavalieri's principle. (fill in options are: r/2- r- 2r- an annulus- a circle -1/3pier^3- 2/3pier^3- 4/3pier^3- 5/3pier^3- 2pier^3- 4pier^3)