project motion problem

denisa

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Oct 6, 2011
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If increasing the initial speed of a projectile by a factor of 5 increases its rane by what factor? Assume the elevation is the same.

Possible answers:
a)factor of 25
b)factor of 5
c)factor of 2.24
d) The elevation must be specified before an answer can be found

I know the Range formula=((v0^2)/g)sin2a
Have problem with this problem... Help me please. Anything will help.. Thanks
 
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Keeping the angle constant, means that sin(2a) remains constant, right?

Range=vo 2sin(2a)g\displaystyle Range = \frac{v_o\ ^2 \sin(2a)}{g}

So, this becomes:

Range=vo 2kg\displaystyle Range = \frac{v_o\ ^2k}{g}

g is also constant, so we can combine those two constants to give another constant. Let's call it c.

Range=cvo 2\displaystyle Range = c v_o\ ^2

You can then set up the proportionality symbol:

Rangevo 2\displaystyle Range \propto v_o\ ^2

Can you work with this one now?
 
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ok is @ means the angle of object? ( which i do not have?) Right?
 
ok, i know that v0 is a speed, but i don't know the speed? if i use 5 as a speed it will give me 25..? Please help me..
 
Not quite the way you do it.

The first 'throw' gives a certain range for a velocity of throw v0:

Rv0 2\displaystyle R \propto v_0\ ^2

This means that:

Rv0 2=k\displaystyle \frac{R}{v_0\ ^2} = k

So, no matter what the velocity or the range, they will still give the constant k.

Now, you increase v0 by a factor of 5. This means, vo5vo\displaystyle v_o \rightarrow 5v_o

The new range, R' is then:

R(5vo)2=k\displaystyle \frac{R'}{(5v_o)^2} = k

But we can write down k as the previous equation!

R(5vo)2=Rvo 2\displaystyle \frac{R'}{(5v_o)^2} = \frac{R}{v_o\ ^2}

Now, you simplify a little:

R25vo 2=Rvo 2\displaystyle \frac{R'}{25v_o\ ^2} = \frac{R}{v_o\ ^2}

R25=R\displaystyle \frac{R'}{25} = R

R=25R\displaystyle R' = 25R

So, the resultant range R' is 25 times the initial range, R
 
I'm glad you do! Feel free to ask any more questions when you have doubts! :)
 
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