Finding when a object hits the ground, its velocity when it hits, etc

[shamieh]

I need someone to check my work, because I am getting very weird numbers. Just want to make sure I am doing this right...
The height of an object in meters after t seconds is given by \(\displaystyle H(t) = 152t - 16t^2\).

a) When does it hit the ground?
My answer: t(152 - 16t) = 0 which means t = 0, \(\displaystyle t = \frac{152}{16}\)

b) What is the velocity when it hits?

152 - 304 = -152 m/s

Is this correct?

 

[MarkFL]

Yes, that's correct...you should expect in this case that the final and initial velocities have the same magnitude, but opposite directions. The negative sign on the final velocity means it is traveling in a downward direction.

 

[HallsofIvy]

I need someone to check my work, because I am getting very weird numbers. Just want to make sure I am doing this right...
The height of an object in meters after t seconds is given by \(\displaystyle H(t) = 152t - 16t^2\).

a) When does it hit the ground?
My answer: t(152 - 16t) = 0 which means t = 0, \(\displaystyle t = \frac{152}{16}\)

Any particular reason why you didn't reduce this to $$\dfrac{19}{2}$$ or 9.5? Also it should be 19/2 seconds or 9.5 seconds.

b) What is the velocity when it hits?

152 - 304 = -152 m/s

Is this correct?

H'= 152- 32t
152- 32(19/2)= 152- 304= -152 m/s

 

[topsquark]

a) When does it hit the ground?
My answer: t(152 - 16t) = 0 which means t = 0, \(\displaystyle t = \frac{152}{16}\)

A picky, but important point. t = 0 s is not an answer here. It is at the ground at t = 0 s, but is being fired upward at that point.

-Dan

 

[MarkFL]

A picky, but important point. t = 0 s is not an answer here. It is at the ground at t = 0 s, but is being fired upward at that point.

-Dan

Good catch, Dan! :D

I don't think that's picky at all. Students should learn early on which roots of their equations apply and which do not and why. (Yes)