calculus
posted by Tatjana .
a woman throws a ball vertically upwards so that 4seconds elapse before it strikes the ground. Determine the maximum height of the ball and the velocity with which the ball was thrown. Assume that the acceleration due to gravity is 10m/s squared and neglect the effect of air resistance

the height h(t) is
h(t) = vt4.9t^2
h=0 when t=4, so
4v4.9*16 = 0
v = 19.6
h(t) = 19.6t  4.9t^2
max h at t=2 because of symmetry
h(2) = 19.6*2  4.9*4 = 19.6