An astronaut notices that a pendulum which took 2.45 s for a complete cycle of swing when the rocket was waiting on the launch pad takes 1.25 s for the same cycle of swing during liftoff.
what is the acceleration (m/s²) of the rocket? (hint: inside the rocket, it appears that g has increased.)

2.84 g's with the remaining 1 g coming from gravity (3.84 g's)

Explanation:

period of oscillation while waiting (T1) = 2.45 s

period of oscillation at liftoff (T2) = 1.25 s

period of a pendulum (T) =2π.

where

therefore the ration of the periods while on ground and at take off will be

=(2π ) /  (2π)

where

=

squaring both sides we have

=

=

assuming that the acceleration on ground a1 = 9.8 m/s^{2}

=

a2 = 9.8 x

substituting the values of T1 and T2 into the above we have

a2 = 9.8 x

a2 = 9.8 x 3.84

take note that 1 g = 9.8 m/s^{2} therefore the above becomes

a2 = 3.84 g's

Hence assuming the rock is still close to the ground during lift off, the acceleration of the rocket would be 2.84 g's with the remaining 1 g coming from gravity.

answer: he got 23

explanation: