Calculating Bullet Height: Moon vs Earth | Time to Reach Maximum Displacement

[PrudensOptimus]

A bullet fired straight up from the moon's surface would reach a height of s = 832t - 2.6t^2 after t sec. On Earth, in the absence of air, its height would be s = 832t - 16t^2 after t sec. How long would it take the bullete to get back down in each case?

On Moon:
It would take the same amount of time to get down as it took to go up(maximum displacement, when v = 0).

Thus, V(t) = ds/dt = 832 - 5.2t = 0. t = 160s. It would take about 2 minutes and 40 seconds.

On Earth:
Same notion. V(t) = ds/dt = 832 - 32t = 0. t = 26s.

-- Am I right?

Question 2:
The position of a body at time t sec is s = t^3 - 6t^2 +9t meters. Find the body's acceleration each time the velocity is 0.

Because V(t) = ds/dt = 3t^2 - 12t + 9,
a(t) = dv/dt = 6t - 12

Particle has v = 0 at t = 3, and 1 sec

Thus, the particle has acelleration at v = 0 at:

a(3) = 6 m/s^2
and
a(1) = -6 m/s^2

-- Am I right?


Please correct my mistake. Thanks.

 

[gnome]

#2 looks right.

#1, a few discrepancies...

First of all, if the position functions are really as you typed them:
s = 832 - 2.6t^2
and
s = 832 - 16t^2

then your derivatives are wrong because there is no t in the first term of each equation.

On the other hand, the problem, as you posted it, doesn't say that s is the MAXIMUM height, and it also isn't clear whether there are two different s's, or only one value of s...

On the other hand, there really should be a t in the first term, since the usual position function is s = v₀t - .5at^2
So should it say s = 832t - 16t^2?

On the other hand, maybe the trick is that the t is "hidden" such that 832 = v₀t and both v₀ and t are unknown...

On the other hand, that acceleration for the moon doesn't really make sense either; free fall acceleration near the surface of the moon is approx. 1.6 m/s^2...

Too many hands.

Bailing out...

 

[einsteinian77]

are you a teacher prudensoptimus?

 

[PrudensOptimus]

Originally posted by einsteinian77
are you a teacher prudensoptimus?

No, I'm a freshman in high school. I'm going to be in 10th grade in beginning of August, so I want to prepare myself for the class.

 

[PrudensOptimus]

Originally posted by gnome

First of all, if the position functions are really as you typed them:
s = 832 - 2.6t^2
and
s = 832 - 16t^2

then your derivatives are wrong because there is no t in the first term of each equation.

I made a typo. Yes there are t's after 832s...

Originally posted by gnome

On the other hand, that acceleration for the moon doesn't really make sense either; free fall acceleration near the surface of the moon is approx. 1.6 m/s^2...B]

I think it makes sense. It takes longer for things to drop on the moon because the acceleration comparing to the Earth's gravity is smaller.

 

[gnome]

OK, then your answer is correct.

(and I just realized that the equations for question 1 are in feet/sec, not meters/sec, so the 2.6t^2 for the moon is OK too)

 

[STAii]

You know, you can make your work really easier (no need for calculus !).
s = 832t - 2.6t^2
s is displacement.
now, displacement will be equal to zero in only two cases (in projectiles) :
1-the object didn't move yet
2-the object went up and back down to the Earth's surface.
So all you have to do is to substitue the value of s with 0, and solve the equation .

And BTW, if i understood ur first way of solving this right, then i think there is something wrong about it. You see the velocity of the bullet will not be 0 when it reaches the surface, it will actually be equal in magnitude (and opposite in direction) of the initial velocity.