Suppose that a ball is thrown directly upward from a height of 2 meters with an initial velocity of 15 m/s. What is the formula for the height(t) of the ball t seconds after it is thrown?
At what time t will the ball in part C hit the ground?
at what time will the ball in part c reach its maximum height? what is this maximum height?
y = 2 + 15 t - 1/2 (9.81)t2
y will equal 0 when the ball hits the ground
0 = 2 + 15t - 1/2 (9.81) t2 solve for t using Quadratic Formula ( throw out negative value)
max height will occur at t = - b/2a = - 15/ (2*4.905) = 15/9.81 = ....... seconds
use this value of 't' in the equation at the top (red) to calculate the max height , y.....
y = 2 + 15 t - 1/2 (9.81)t2
y will equal 0 when the ball hits the ground
0 = 2 + 15t - 1/2 (9.81) t2 solve for t using Quadratic Formula ( throw out negative value)
max height will occur at t = - b/2a = - 15/ (2*4.905) = 15/9.81 = ....... seconds
use this value of 't' in the equation at the top (red) to calculate the max height , y.....