Solving Velocity Problems on the Moon

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The discussion focuses on solving a velocity problem related to an arrow shot upward on the Moon, described by the height equation H=58t-.83t². The velocity after one second has been calculated, and the method to find the velocity at time t=a is explained as plugging a into the velocity equation, yielding 58a-.83a². To determine when the arrow hits the Moon, the equation is set to zero, resulting in a time of approximately 69.88 seconds. The impact velocity is calculated by substituting this time back into the velocity equation, resulting in 1957.04 m/s. The discussion emphasizes careful reading of the problem and proper application of equations to find unknowns.
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Calc help - velocity problem...

Just looking to be pointed in the right direction, for now. I understand the equations I have learned about velocity, but am having trouble setting up this word problem:

If an arrow is shot upward on the moon, with a velocity of 58m/s, it's heigt in meters after t seconds is given by
H=58t-.83t^{2}

Find the velocity of the arrow after one second? (I already figured this out)
Find the velocity of the arrow when t=a?
When will the arrow hit the moon? (I already figured this out)
With what velocity will the arrow hit the moon?
 
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Set H = 0, solve for t then find the corresponding velocity.

Or, since energy is conserved and there is no air resistance then the kinetic energy will be the same at the end of the trajectory as it was at the beginning of the trajectory since the potential energy has a single value at the surface - therefore the speed is the same but the arrow is headed downward.
 


To find the velocity of the arrow when t=a, we can simply plug in a for t in the equation H=58t-.83t^{2}. So, the velocity at t=a would be 58a-.83a^{2}.

To find when the arrow will hit the moon, we can set the equation equal to 0 and solve for t. So, 58t-.83t^{2}=0. This gives us two solutions, t=0 and t=69.88 seconds. Since t=0 represents the initial position, we can disregard it and say that the arrow will hit the moon after approximately 69.88 seconds.

To find the velocity at which the arrow hits the moon, we can plug t=69.88 into the equation for velocity, which would give us a velocity of 58(69.88)-.83(69.88)^{2}=1957.04 m/s.

I hope this helps point you in the right direction and gives you a better understanding of how to approach velocity problems. Remember to always carefully read the problem and identify what information you have and what you need to find. Then, use the appropriate equations and plug in the given values to solve for the unknown variable. Good luck with your calculations!
 
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