A ball is thrown upward from a height of 5.6 m at an initial speed of 49 m/sec. Acceleration resulting from gravity is − 9.8 m/ sec^2

Question

 Neglecting air resistance, solve for the velocity  and the height  of the ball  seconds after is is thrown.

Find the velocity (meter/second)and the hight (meters)

Answer 1

h(t) = -4.9t^2 + 49t + 5.6

v(t) = h'(t) = -9.8t + 49

a(t) = v'(t) = h"(t) = -9.8