Arocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8 m/s^2. the acceleration period lasts for time 8.00 s until the fuel is exhausted. after that, the rocket is in free fall. find the maximum height ymax reached by the rocket. ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 m/s^2 .

The maximum height will be 7408.8 meters

Explanation:

final velocity = initial velocity + acceleration × time  

final velocity = 0 m/s + 58.8 m/s^2 ×6 s  

Final velocity = 352.8 m/s  

final velocity ^2 = initial velocity ^2 + 2 × acceleration × displacement  

(352.8)^2 = (0)^2 + 2×58.8 ×displacement  

Solving for displacement,

height = 1058.4 meters.  

After this, the rocket is in free fall, we can use the same equation.  

final velocity ^2 = initial velocity ^2 + 2 ×acceleration×displacement  

final velocity = 0  

0^2 = 352.8^2 + 2×(-9.8)×displacement  

displacement = 6350.4 meters  

the maximum height will be 7408.8 meters

atomic = no of protons.

mass number-atomic number = neutron number

i guess it’s a and b