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

  • Thread starter PrudensOptimus
  • Start date
In summary: So, if u used the first method, you should check when the velocity is equal in magnitude to the initial velocity (not when it is 0).In summary, to find out how long it would take for a bullet fired straight up from the moon's surface to reach its maximum height of s = 832t - 2.6t^2 after t seconds, we can set the displacement to 0 and solve the equation. This would give a time of approximately 2 minutes and 40 seconds. On Earth, in the absence of air, the bullet's maximum height would be s = 832t - 16t^2 after t seconds, and using the same method, we can find that it would take approximately
  • #1
PrudensOptimus
641
0
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.
 
Last edited:
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  • #2
#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...
 
  • #3
are you a teacher prudensoptimus?
 
  • #4
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.
 
  • #5
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.
 
Last edited:
  • #6
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)
 
  • #7
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 :smile:.

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.
 

1. What is calculus?

Calculus is a branch of mathematics that deals with rates of change and the accumulation of quantities. It is used to solve problems in physics, engineering, economics, and many other fields.

2. What are the two main branches of calculus?

The two main branches of calculus are differential calculus and integral calculus. Differential calculus focuses on the rate of change of a function, while integral calculus deals with the accumulation of quantities.

3. What is the difference between derivatives and integrals?

Derivatives and integrals are two fundamental concepts in calculus. Derivatives measure the rate of change of a function at a specific point, while integrals measure the accumulation of a quantity over a given interval.

4. How is calculus used in real life?

Calculus has many real-world applications, including predicting the motion of objects, optimizing systems, and analyzing data in fields such as physics, engineering, economics, and biology.

5. What are some common methods for solving calculus problems?

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