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?

(impact)

(impact)

Explanation:

The equation for height is

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

So we have (adding units at the end):

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

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

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

The velocity at the time it hits the moon:

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

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your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

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