A rocket is fired vertically with an acceleration of 16m/s^2. After time T, the rocket's engine runs out of fuel and it begins to fall freely. The maximum height of the rocket is 940m, what is the minimum value of T?

the distance traveled by the rocket in T seconds with the acceleration 16m/s.s is=940m

velocity=16*Tm/s=16Tm/s
during T seconds the rocket moved upward with velocity 16T m/s and downward accelaration due to gravity till its velocity becomes zero.
0=(16T*16T)-2gh2
= 256T.T-2*10h2
h2=256T.T/20
now,256T.T+160T.T/20=940
or,416T.T=940
or,T.T= 940/416=2.25
or, T=1.5s

To find the minimum value of T, we need to determine the time it takes for the rocket to reach its maximum height of 940m.

We know that the acceleration of the rocket while it is firing is 16m/s^2, and the rocket is following a vertical motion. We can use the kinematic equation for vertical motion:

h = ut + (1/2)at^2

Where:
h is the height
u is the initial velocity (which is 0 since the rocket starts from rest)
a is the acceleration
t is the time

Plugging in the known values, we have:

940 = 0*t + (1/2)*16*t^2

Simplifying the equation, we have:

940 = 8t^2

Dividing both sides by 8:

t^2 = 940/8

t^2 = 117.5

Taking the square root of both sides:

t = √117.5

t ≈ 10.83 seconds

Therefore, the minimum value of T is approximately 10.83 seconds.