Answer:
The velocity of the arrow after 3 seconds is 30.02 m/s.
Step-by-step explanation:
It is given that,
An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:
![h(t)=35t-0.83t^2](https://img.qammunity.org/2021/formulas/physics/high-school/kf49t8morejel6a15gtdmj6fra7e7sa6cs.png)
We know that the rate of change of displacement is equal to the velocity of an object.
![v(t)=(dh(t))/(dt)\\\\v(t)=(d(35t-0.83t^2))/(dt)\\\\v(t)=35-1.66t](https://img.qammunity.org/2021/formulas/physics/high-school/ljx91l6svucvq7zvs0ty0q4fyxgqgbpkta.png)
Velocity of the arrow after 3 seconds will be :
![v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s](https://img.qammunity.org/2021/formulas/physics/high-school/39zth6uen3qvm398md4k30r16ryl915u0b.png)
So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.