Basic Physics

[Logic-cake]

My teacher hasn't explained much, and I'm not sure how to get the answer to these problems.

 

If a projectile is fired straight up at a speed of 30 m/s, the total time to return to its starting point is about___.

 

Supposedly the answer is 6 seconds, but I don't see how. Wouldn't I divide 30 m/s by the acceleration of gravity to get about 3 seconds?

 

Also, I don't understand the answer to this one:

The vertical height attained by a basketball player who achieves a hang time of a full one second is about___.

 

and the answer is supposedly 1.2 meters. I tried to substitute the values into the equation [math] d= \frac {1}{2}gt^{2}[/math], where g is approx. [math]10 m/s^{2}[/math] and [math]t[/math], the hang time, is 1 second, and got the value 5. Am I doing something wrong, or is this right?

^{And one more.}

^{If a projectile is fired beneath the water, straight up, and breaks through the surace at a speed of 13 m/s, to what height above the water will it ascend?}

Do I even have enough information to solve this one? Would I not need to know how deep the object is under water before it's fired?

 

Thanks in advance for your help.

[DJBruce]

If a projectile is fired straight up at a speed of 30 m/s, the total time to return to its starting point is about___.

 

Supposedly the answer is 6 seconds, but I don't see how. Wouldn't I divide 30 m/s by the acceleration of gravity to get about 3 seconds?

 

You will want to use the equation:

 

[math]d=\frac{1}{2}gt^{2}+v_{0}[/math]

 

What would the displacement be when the projectile returns to the starting location? Once you answer this use that as d, and then simply solve the equation for t.

 

 

The vertical height attained by a basketball player who achieves a hang time of a full one second is about___.

 

and the answer is supposedly 1.2 meters. I tried to substitute the values into the equation [math] d= \frac {1}{2}gt^{2}[/math], where g is approx. [math]10 m/s^{2}[/math] and [math]t[/math], the hang time, is 1 second, and got the value 5. Am I doing something wrong, or is this right?

 

One second is how long the object is in the air total, however, for half of that time it is on the decesent so therefore you should use .5 s instead.

 

^{And one more.}

^{If a projectile is fired beneath the water, straight up, and breaks through the surace at a speed of 13 m/s, to what height above the water will it ascend?}

Do I even have enough information to solve this one? Would I not need to know how deep the object is under water before it's fired?

 

Yes you can solve this. You have [math]v_0[/math], [math]a[/math]. So which equation would give you displacement given those two variables?

[Mr Skeptic]

Another thing, when working in science always always always write the units in your calculations. Units can be canceled or replaced with equivalent number of other units, just like if they were a variable or unknown like x or y. Had you done this, you'd have immediately noticed you made a mistake (your answer would have the wrong units), and the same will be true later on in various scientific areas. In many cases (but not this one) the units you start with and units you should end with are all you need to know to solve the problem.

 

You will want to use the equation:

 

[math]d=\frac{1}{2}gt^{2}+v_{0}[/math]

 

He means

[math]d=\frac{1}{2}gt^{2}+v_{0}t[/math]

 

and also don't forget that both g and [math]v_{0}[/math] have a direction signified by their sign.

 

^{And one more.}

^{If a projectile is fired beneath the water, straight up, and breaks through the surace at a speed of 13 m/s, to what height above the water will it ascend?}

Do I even have enough information to solve this one? Would I not need to know how deep the object is under water before it's fired?

 

Sometimes they give you irrelevant information. Ignore the irrelevant information.

[DJBruce]

He means

[math]d=\frac{1}{2}gt^{2}+v_{0}t[/math]

 

Ahh, thanks a lot. I must have missed that. +1 for fixing my mistake.

[swansont]

 

Sometimes they give you irrelevant information. Ignore the irrelevant information.

 

The ability to recognize the information as irrelevant is a test of your mastery of the material. It will happen on tests, too, so get used to it.