Question

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

Answer 1

To solve this problem, we will need the standard kinematics equation
$distance=v_0t+\frac{1}{2}at^2$
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)

$distance=v_0t+\frac{1}{2}at^2$
$0=42t+\frac{1}{2}(9.81)t^2$
factor and solve
$t(42+\frac{1}{2}(-9.81)t)=0$
=>
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)