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T= 3.34
Vi= 0
A= 9.81
D= ?
d=Vit+1/2at^2
d= 1/2(9.81)(3.34)2
d= 54.7 or 55 meters tall
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Explanation:
It is given that, the position of a particle as as function of time t is given by :

Let v is the velocity of the particle. Velocity of an object is given by :

![v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%5B%288t%2B9%29i%2B%282t%5E2-8%29j%2B6tk%5D%7D%7Bdt%7D)

So, the above equation is the velocity vector.
Let a is the acceleration of the particle. Acceleration of an object is given by :

![a=\dfrac{d[8i+4tj+6k]}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5B8i%2B4tj%2B6k%5D%7D%7Bdt%7D)

At t = 0, 

Hence, this is the required solution.
Answer:
(a) A = m/s^3, B = m/s.
(b) dx/dt = m/s.
Explanation:
(a)

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.
(b) 
This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.