Arocket initially at rest acceleration at a rate of 99.0meter/second^2. calculate the distance covered by the rocket if it attains a final velocity of 445 meters/seconds after 4.50 seconds.

Distance traveled by the rocket= 1002 m

Explanation:

initial velocity= Vi=0

final velocity= V=445 m/s

time=4.5 s

Acceleration=a=99 m/s²

using kinematic equation

h= Vi t + 1/2 at²

h= 0(4.5) + 1/2 (99) (4.5)²

h=1002 m

Thus the distance traveled by the rocket= 1002 m

the answer is b ) square root of 5

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.