physics anyone?

Question

i'm soo stuck.. thanx for those who help out =]
Two tanks are engaged in a training exercise on level ground. The first tank fires a paint-filled training round with a muzzle speed of 257 m/s at an angle 15.0 degrees above the horizontal while advancing toward the second tank with a speed of 13.5 m/s relative to the ground. The second tank is retreating at a speed of 32.0 m/s relative to the ground, but is hit by the shell. You can ignore air resistance and assume the shell hits at the same height above ground from which it was fired.

1. Find the distance between the tanks when the round was first fired.
Take free fall acceleration to be g = 9.80 m/s^2.

2. Find the distance between the tanks at the time of impact.
Take free fall acceleration to be g = 9.80 m/s^2.

Answer 1

I answered a similar question question before
Read the answer and try to comprehend. Please tell me where you have trouble understanding. email me @ [e-mail] or just add details to the question.


http://answers.yahoo.com/question/index;_ylt=AmK_ng8lYfmx1PrceI_iS1Psy6IX?qid=20070417002407AAvPdVx&show=7#profile-info-XonMrp8kaa

Answer 2

Treat the horizontal and vertical components of velocity separately. First find how long the shell remains in the air, and then use that time to compute how far the shell and the tanks have traveled.

The vertical component of initial velocity vy0 = 257*sin(15º). The vertical velocity is vy(t) = vy0 - g*t. The shell reaches maximum height when vy(t) = 0, or t = vy0/g. It takes the same time to come down, so the total time aloft t = 2*vy0/g = 2*257*sin(15º)/g.

During that time, the firing tank moved 13.5*t, and the target tank moved 32.0*t. The distance the shell traveled is vx*t, where vx = horizontal component of shell velocity, which is 257*cos(15º) + 13.5; so the shell travels [257*cos(15º) + 13.5]*t .

The shell will catch up to the target tank when the shell travels a distance d0 + 32*t, where d0 is the initial distance between the tanks.

257*cos(15º) + 13.5]*t = d0 + 32*t

where t = 2*257*sin(15º)/g. Solve for d0

The distance between the tanks is d0 - 13.5*t + 32.0*t. Use d0 and t as computed above.

Answer 3

I'm a bit rusty on these problems, but lets see how I go...

First thing to work out is how long is the projectile in the air. (I think in a frictionless environment you can say that it takes the same time to go up as it does to comes down.)

The projectile is fired at 257 m/s at an angle of 15 o.
Draw a right angled triangle that converts this information into a vertical and horizontal component and work out the vertical component

sin 15 = u (vert) / 257
u(vert) = sin 15 * 257
u (vert) = 66.5 m/s

Considering the vertical component of the projectile's flight, it goes up and reaches a maximum height at which the velocity is 0 (just before it starts to fall again)

The time it reaches the top of the flight path can be calculated by
v = u + at rearranged for
t = (v - u) / a
v = 0 (final velocity), u = 66.5 m/s and a = -9.8 m/ss (its deccelerating)

t = (0-66.5) / -9.8 = 6.787 s

I think the trick of this is that this is half the flight time (as the projectile hits at the same height that it was fired from, and the atmosphere is frictionless).

So the total flight time of the projectile is 13.575 seconds.

Now claculate how far the projectile travelled in the horizontal direction. The question says that the projectile was fired at 257 m/s muzzle speed, BUT the muzzle was attached to (and travelling with) the tank at 13.5 m/s at the time of firing.

The initial horizontal velocity of the projectile is then

u (horizontal) = (cos 15 * 257) PLUS 13.5 m/s
u (horiz) = 248.2 + 13.5 = 261.7 m/s

Calculate displacement using
s = ut + 1/2att
u = 261.7, t = 13.575
note that a = 0 !! the projectile is travelling at constant velocity in the horizontal plane.

so, s (projectile) = ut = 3553 m

Now, draw a line that shows the projectile travelling 3553 metres horizontally in 13.575 seconds. The left extremetity of that line represents the location of tank 1 at t=0 and the right extremity of the line represents the location of tank 2 at t = 13.575 seconds

We have to calculate the position of tank 1 at t = 13.575 seconds and the position of tank 2 at t=0... clear as mud isn't it! remembering that the line is 3553 metres long.

Tank 1 first:
at t = 0 tank 1 is at s = 0

tank is travelling at u = 13.5 m/s
at t = 13.575
s = ut = 13.5 * 13.575 = 183.3 metres

Now for tank 2:
at t = 13.575 tank 2 was at position 3553 metres
tank 2 had been travelling at 32 m/s before being blown to pieces...

s = ut = 32 * 13.575 = 434.4 metres. This is the distance travelled from t = 0 to t = 13.575. Therefore
at t = 0 tank 2 was at position 3553 - 434.4 metres = 3118.6 metres.

In summary:
at t = 0,
tank 1 was at s = 0 and tank 2 was at s = 3118.6 m
They were 3118.6 metres apart when the round was first fired.

at t = 13.575
tank 1 was at position s = 183.3 metres and
tank 2 was at position s = 3553 metres
They were 3369.7 metres apart at the time of impact.


*phew* that was kind of fun!
If my answers are different to others one reason will be that I have added 13.5 m/s to the horizontal velocity of the projectile because it is being carried by the tank....

Answer 4

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Answer 5

1. 3117.6 m

2. 3368.6 m