Question

If an arrow is shot straight upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by s(t)=58t−0.83t2. What is the velocity of the arrow (in m/s) after 7 seconds? equation editorEquation Editor After how many seconds will the arrow hit the moon? With what velocity (in m/s) will the arrow hit the moon?

Answer 1

a)
`v(7s)=46.38m/s`

b)
`t=70s`

c)
`v(70s)=-58.2m/s`

Step-by-step explanation:

From the exercise we know the arrow's equation of position

`s(t)=58t-0.83t^2`

a) If we want to know the velocity of the arrow after t=7s we need to derivate the equation of position to get velocity

`v(t)=(ds)/(dt)=58-1.66t`

Now, we evaluate 7s in the equation

`v(7s)=58m/s-(1.66m/s^2)(7s)=46.38m/s`

b) To find how much time does it take the arrow to hit the moon we need to evaluate s=0 at the equation of position in the vertical direction

`s(t)=58t-0.83t^2`

`0=58t-0.83t^2`

Solving the quadratic equation using the following formula

`t=(-b±√(b^2-4ac))/(2a)`

`a=-0.83\\b=58\\c=0`

`t=0` or
`t=70s`

Since time can not be zero the answer is t=70s

c) To calculate the arrow's velocity when it hits the moon we need to evaluate t=70s in the equation of velocity

`v(70s)=58m/s-(1.66m/s^2)(70s)=-58.2m/s`