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? After how many seconds will the arrow hit the moon? With what velocity (in m/s) will the arrow hit the moon?

Answer 1

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

`t(0m)=69.88s` (impact)

`v(0m)=-58 m/s` (impact)

Step-by-step explanation:

The equation for height is
`s(t)=58t-0.83t^2`

We derive the equations for velocity and acceleration by derivating it respect with time:

`v(t)=(ds(t))/(dt)=58-2(0.83)t=58-1.66t`

`a(t)=(dv(t))/(dt)=-1.66`

So we have (adding units at the end):

The velocity after 7s (evaluating the velocity formula for 7):

`v(7)=58-1.66(7)`

`v=46.38 m/s`

The time it hits the moon (when the height is null again):

`s(t)=0=58t-0.83t^2`

One solution is t=0, which corresponds to departure of course, but we want the other solution, which is for impact.

`0=58-0.83t`

`0.83t=58`

`t=69.88s`

The velocity at the time it hits the moon:

`v(69.88)=58-1.66(69.88)`

`v=-58 m/s`

As it should be since the arrow left the moon at 58m/s.