To solve this problem, we will need the standard kinematics equation
![distance=v_0t+\frac{1}{2}at^2](https://tex.z-dn.net/?f=distance%3Dv_0t%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
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](https://tex.z-dn.net/?f=distance%3Dv_0t%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![0=42t+\frac{1}{2}(9.81)t^2](https://tex.z-dn.net/?f=0%3D42t%2B%5Cfrac%7B1%7D%7B2%7D%289.81%29t%5E2)
factor and solve
![t(42+\frac{1}{2}(-9.81)t)=0](https://tex.z-dn.net/?f=t%2842%2B%5Cfrac%7B1%7D%7B2%7D%28-9.81%29t%29%3D0)
=>
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)
Since you are solving for a, you want to have a on one side of the equation and the other terms on another side of the equation. It would be easiest to have all the terms with a on the left side of the equation, so that is what we will do.
Subtract 9a from both sides to get a on the left side of the equation.
5 + 5a = -5
Subtract 5 from both sides of the equation to isolate the term with a.
5a = -10
Divide both sides of the equation by 5 to solve for a.
a = -2
Answer: move 7 places to the right and 3 places down
Step-by-step explanation:
Answer:
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Step-by-step explanation: