It should decellerate with -362167 m/s^2.

Basically Delta V/T:

Delta V is 412 - 330

T = =0,084/ ((330+412)/2)

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This is a discussion on *Projectiles* within the **Members' Club Room** forums, part of the China Defense & Military category; Ppl i need help with an engineering question: Need help with getting the answer. A bullet is fired through a ...

- 03-28-2007, 03:12 AM #1
## Projectiles

Ppl i need help with an engineering question: Need help with getting the answer. A bullet is fired through a board, 8.4 cm thick, with its line of motion perpendicular to the face of the board. If it enters with a speed of 412 m/s and emerges with a speed of 330 m/s, what is the bullet's acceleration as it passes through the board?

any help will be good

- 03-28-2007, 03:50 AM #2Member
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## Re: Projectiles

It should decellerate with -362167 m/s^2.

Basically Delta V/T:

Delta V is 412 - 330

T = =0,084/ ((330+412)/2)

- 03-28-2007, 12:46 PM #3Senior Member
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## Re: Projectiles

[note: s is the distance, don't know correct figure in english here, v is "delta v" or 82m/s]

I disagree: s = 1/2*a*t²; you have T = 2,264*10^-4s

If you insert that in the above equiation, you get s = 9,28*10^-3m, wich is a little more than 1/10 of the original given 0,084m thickness.

My solution:

s = 1/2*a*t^² (1); and v = a*t (2); (1) in (2) gives you s = 1/2*a*t (3)

=> from (3): t = s / (1/2*v) = 0,084 / (0,5*412-330) = 2,05*10^-3s

=> now in (1)**a**= s / (0,5*t² =**39976 m/s²**

check: insert results in (1)**s**= 0,5*39976*(2,05*10^-3)² =**0,084m**

Skorzeny: I think your mistake is "T = =0,084/ ((330+412)/2)" you are assuming an average velocity of the projectile in the board wich would need a linear decceleration over distance, but it is quadratic. Note: s=1/2*a*t**²**

- 03-28-2007, 01:28 PM #4Member
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## Re: Projectiles

Hi.

You forgot a part of your equation, it should be

s = V(initial)*T + 1/2*a*T^2

(Because it is decellerating, 1/2*a*T^2 is negative.)

That confirms my answer. And yes, the decelleration is not quadratic.

What you should have seen in you reply, is that your estimate for time is far higher than what you would get with the terminal speed.

But nice try

- 03-28-2007, 01:48 PM #5Senior Member
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## Re: Projectiles

Ok, you proofed me wrong. (I didn't check but it sounds completely logic) I accept my inferiority

- 03-29-2007, 03:04 AM #6New Member
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## Re: Projectiles

hey i was wondering if you guys would help me with these problems there for a physics assignment. please help.

1: Starting from rest, a car travels 1,400 meters in 2.0 minutes. It accelerated at 1.2 until it reached its cruising speed. Then it drove the remaining distance at constant velocity. What was its cruising speed?

2:A tennis player standing 10.9 m from the net hits the ball at 4.00 above the horizontal. To clear the net, the ball must rise at least 0.370 m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racket?

i know these are not related to china's army but you guys seem to know wat to do.

thanks for your time

- 03-29-2007, 07:07 AM #7
## Re: Projectiles

thanx ppl for the help, much appreciate it

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