Velocity Problem: If an arrow is shot straight upward on the

[needurhelp]

If an arrow is shot straight upward on the moon with a velocity of 79 m/s, its height (in meters) after t seconds is given by h = 79t - 0.83t^2 .
What is the velocity of the arrow (in m/s) after 5 seconds?
After how many seconds will the arrow hit the moon?
With what velocity (in m/s) will the arrow hit the moon?

i was able to find the part first. took the derivative of h = 79t-0.83t^2
and got 79-1.66x, pluged 5 in for x. and got 70.4
im not sure how do to the other parts, can someone help? thnx

 

[tkhunny]

You just need to solve h = 0 for t. Isn't that when the arrow comes back down?

Don't forget to move before it returns.

 

[needurhelp]

tkhunny, im not sure what u mean by

You just need to solve h = 0 for t.

 

[skeeter]

tkhunny, im not sure what u mean

he means ...

79t - 0.83t² = 0

solve for t.

use your velocity expression with the time found above to answer the last question.