An expression for the velocity can be obtained by differentiating the distance term with respect to time. Therefore
dH/dt = v = 10 - 3.72t
substituting t = 2 sec
v = 10 - 3.72(2)
v = 2.56 m/s, velocity after 2 hours
Since there is no variable x in the problem, I assumed that x = t =a
therefore at t = a
v = 10 - 3.72a^2
when the rock hits the surface:
when the rock reaches the top, the velocity is equal to zero
therefore the expression becomes, H = 1.86t2
dH/dt = v = 3.72t
when the rock hits the surface H = 10 therefore t = 2.32 s
@t = 2.32 s
v = 3.72*2.32 = 8.63 m/s
