it is well known that the equations are
s(t) = 3000+75t-16t^2
v(t) = 75-32t
So plug and chug
Extra credit: derive those equations.
Near the surface of the earth, the acceleration of a falling body due to gravity is 32 feet per second per second, provided that air resistance is neglected. If an object is thrown upward from and initial height of 3000 feet with a velocity of 75 feet per second, find its velocity and height 6 seconds later.
Velocity =
Height =
2 answers
hint
if h = height
d^2h/dt^2 = acceleration = -32 ft/s^2 constant
dh/dt = speed up
if h = height
d^2h/dt^2 = acceleration = -32 ft/s^2 constant
dh/dt = speed up