Magnitude of Acceleration Question

[Doug Desatnik]

A projectile is fired with an initial speed of 50.0 m/s. Neglect air
resistance. What is the magnitude of the acceleration of the projectile
3.00 sec after it is fired?

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I am using the following equation:

A=(V-Vo)/(T-To)

I set up the variables in the following way:

a = ?
Vo = 0 m/s
V = 50.0 m/s
To = 0 sec
T = 3 sec

When I do the math, the answer comes out to be 16.6 m/s. Can anybody verify that I did this problem correctly? If not then can anybody let me know where I might have gone wrong and pioint me toward the right track?

Thanks!

Doug

 

[Tide]

The only force acting on the projectile is gravity so the acceleration should be g. Are you sure you stated the problem correctly?

 

[pervect]

You're off on the wrong track, I'm afraid.

Here's a simpler problem that may get you on the right track:

Drop an object from some height, h. When you release it, what's the acceleration of the object, neglecting air resistance?

1 second into the fall, what's the acceleration of the object?

2 seconds into the fall, what's the accelerationof the object?

 

[Doc Al]

Welcome to PF!

Think about it. What's the acceleration of a projectile at any time if air resistance is ignored?

Does that acceleration depend on the speed?

 

[Rogerio]

It seems there is only the gravity acceleration (10m/s^2). It is constant and doesn't depend on the speed.

 

[Doug Desatnik]

Thank you all for your replies!

Here is the complete question
"A projectile is fired horizontally with an initial speed of 50.0 m/s. Neglect air resistance. What is the magnitude of the acceleration of the projectile 3.00 s after it is fired?"

Doc Al, wouldn't the acceleration of a projectile at any time if air resistance is ignored be constant, meaning that if air resistence is ignored the particle would never slow down right?

pervect, neglecting air resistence, the speed of the object would be the same at both 1 and 2 seconds right? Is saying "without air resistence" in effect referring to a vacuum?

Rogerio, How did you come up with 10 m/s^2?

- Doug

 

[Doc Al]

Doc Al, wouldn't the acceleration of a projectile at any time if air resistance is ignored be constant

Yes, the acceleration is constant: acceleration due to gravity = g downwards, which is approximately 9.8 m/s^2 directed down.

, meaning that if air resistence is ignored the particle would never slow down right?

No, the acceleration is constant, not the velocity! Commonsense should tell you that what is thrown up will come down.

 

[Doug Desatnik]

Yeah - what must come up MUST of course come down :) I get the difference between velocit and acceleration now.

So then to solve the problem at hand, I figure I need to use the laws of kinematics...

If so, the I set the variables up like so:

x = ? (NEED TO SOLVE FOR)
a = -9.80 ,/s^2
V(initial) = 50.0 m/s
V(final) = ?
t = 3 sec.

But I don't see a kinematics eauqtion to solve for X when presented with the a, V(initial) and time variables. All of these equations seem to need the V(final)??

Am I hopelessly lost on this or is there hope!? Are the kinematics equations what I need?

Thank!

- Doug

 

[arildno]

It's a trick question, Doug.
Students are meant to fall into this trap.
The acceleration is, and remains "g", irrespective the information you've been given concerning initial velocity.
hence, you don't need any equations to solve the problem properly.

 

[Doug Desatnik]

Ah-HA! Now I get it. It's like the light just went off after reading all of these replies. Now it makes total sense.

Thank you all for your help! I look forward to posting again and eventually being able to help others as well.

- Doug