Arocket is launched vertically from the ground with an initial velocity of 56ft/sec. write a quadratic function h(t) that shows the height, in feet, of the rocket t seconds after it was launched.

h(t) = 56*t  - 16.1*t^2

Step-by-step explanation:

The general quadratic formula for these cases is

h(t) = h0 + v*t + 0.5*a*t^2

Where

a = the acceleration of the body. In this case, since the rocket is going upwards, we take the acceleration of the gravity.

a = -32.2 feet/s^2 = -9.8 m/s^2

v = the velocity

v = 56 ft/s

h0 =  the initial distance. Since it launched from ground

h0 = 0

h(t) = 56*t - 16.1*t^2

See, attached picture

6: 07 is your answer

step-by-step explanation: