Aflying cannonball’s height is described by formula y=−16t^2+300t. find the highest point of its trajectory. in how many seconds after the shot will cannonball be at the highest point?

Highest point = 1406.25 Number of seconds = 9.375 We've been given the quadratic equation y = -16t^2 + 300t which describes a parabola. Since a parabola is a symmetric curve, the highest value will have a t value midway between its roots. So using the quadratic formula with A = -16, B = 300, C = 0. We get the roots of t = 0, and t = 18.75. The midpoint will be (0 + 18.75)/2 = 9.375 So let's calculate the height at t = 9.375. y = -16t^2 + 300t y = -16(9.375)^2 + 300(9.375) y = -16(87.890625) + 300(9.375) y = -1406.25 + 2812.5 y = 1406.25 So the highest point will be 1406.25 after 9.375 seconds. Let's verify that. I'll use the value of (9.375 + e) for the time and substitute that into the height equation and see what I get.' y = -16t^2 + 300t y = -16(9.375 + e)^2 + 300(9.375 + e) y = -16(87.890625 + 18.75e + e^2) + 300(9.375 + e) y = -1406.25 - 300e - 16e^2 + 2812.5 + 300e y = 1406.25 - 16e^2 Notice that the only term with e is -16e^2. Any non-zero value for e will cause that term to be negative and reduce the total value of the equation. Therefore any time value other than 9.375 will result in a lower height of the cannon ball. So 9.375 is the correct time and 1406.25 is the correct height.

Num of plants = (original#ofplants)*(1.6)^t