The initial vertical velocity of a rocket shot straight up is 42 meters per second. how long does it take for the rocket to return to earth? round your answer to the nearest tenth of a second

To solve this problem, we will need the standard kinematics equation

where
v0=initial velocity (positive upwards) m/s
t=time in seconds
a=acceleration due to gravity (positive upwards) = -9.81 m/ ²

Substituting values,
distance = 0 (when it falls back on earth)
v0=42 m/s
a=-9.81 (negative means towards earth, downwards)



factor and solve

=>
t=0 (beginning of launch), or
42-(1/2)*9.81t=0 => t=2*42/9.81=8.563 s. 

Time to return = 8.6 seconds (to the nearest tenth of a second)

(m+11n)^2=(m+11n)(m+11n)=m^2+11mn+11mn+121n^2=m^2+22mn×121n^2 for example; use foil: (a+b)(a+b)first. a^2outside. abinside. ablast. b^2