Computers and Technology: Ridges help us objects.

To solve this problem, we will need the standard kinematics equationwherev0=initial velocity (positive upwards) m/st=time in secondsa=acceleration due to gravity (positive upwards) = -9.81 m/ ²Substituting values,distance = 0 (when it falls back on earth)v0=42 m/sa=-9.81 (negative means towards earth, downwards)factor and solve= t=0 (beginning of launch), or42-(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)...

Ridges help us objects. - id9145730, carolin7493

Ответ:

alazhamason

20.01.2020

Computers and Technology

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)

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