An object is shot upwards from ground level with an initial velocity of 3 meters per second; it is subject only to the force of gravity (no air resistance).

Question

Find its maximum altitude and the time at which it hits the ground.

Answer 1

h(t) = -4.9t^2 +3t

h'(t) =-9.8t +3 = 0

t = 3/9.8 = about 0.306 = time in seconds when the object reaches maximums height. Plug that value into the original equation to solve for maximum height

h(3/9.8) = -4.9(3/9.8)^2 +3(3/9.8) = -0.459 + 0.918 = 0.459 meters = maximum height

the object hits ground level when h(t) = 0 = -4.9t^2 +3t = t(-4.9t+3)

t=0 and t =3/4.9 = about .612 seconds= twice the time it takes to reach the maximum height: 2(.306) = 0.612

plug 3/4.9 into h(3/4.9) = -4.9(3/4.9)^2 +3(3/4.9) = 0= ground level

the h(t) equation is (a/2)t^2 + vt where a= -9.8 meters per second per second, the deceleration due to gravity on earth at sea level. v= initial velocity