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the super spy james bond is riding on his super bike at a uniform speed of 60 kmph towards a rocket launcher in chinese territory.the road is straight and horizontal.when he reaches the point x on his way he look at the top of the rocket launcher and finds the angle of elevation to be 30 degree. after 1min. and 30 sec.he looks at the top of the rocket launcher again and finds the angle of elavation is now increased by 30 degree. he takes out a powerful bomb from his pocket and thinks how many sec. he has now,to reach the base of the rocket launcher

can you help him and tell him how many sec. he has?

 Oct 19, 2015

Best Answer 

 #2
avatar+118609 
+10

Hi Manas, it is nice to hve your on the forum  laugh

This is a great question :))     

 

the super spy james bond is riding on his super bike at a uniform speed of 60 kmph towards a rocket launcher in chinese territory.the road is straight and horizontal.when he reaches the point x on his way he look at the top of the rocket launcher and finds the angle of elevation to be 30 degree. after 1min. and 30 sec.he looks at the top of the rocket launcher again and finds the angle of elavation is now increased by 30 degree. he takes out a powerful bomb from his pocket and thinks how many sec. he has now,to reach the base of the rocket launcher

can you help him and tell him how many sec. he has?

 

He is travelling at  60mph

This converts to  1000m/s

 

When he starts thinking, which is instantaneous becsues he IS James Bond,  let him be y metres from his target.

Ther the height of the launcher be h

 

In 1 minute and 30 seconds he will have travelled  1.5*1000=1500 metres

so at point X he will be  1500+y  metres frpom the base of the launcer.

 

so

\(tan30=\frac{h}{y+1500} \qquad and \qquad tan60=\frac{h}{y}\\ \frac{1}{\sqrt3}=\frac{h}{y+1500} \qquad and \qquad \frac{\sqrt3}{1}=\frac{h}{y}\\ \frac{y+1500}{\sqrt3}=h \qquad and \qquad \sqrt3\;y=h\\ \frac{y+1500}{\sqrt3} =\sqrt3\;y\\ y+1500 =3\;y\\ 1500 =2\;y\\ y=750 metres\\ \)

 

\(750m\times\frac{1min}{1000m}=0.75 minutes = 45seconds\)

 

 

If conditions stay perfect he has 45 seconds to go.

 

If you have a problem understanding then let me know but make sure you draw the pic and try to work it out first.  :)

 Oct 20, 2015
 #1
avatar
+1

No, I can't.  I need either more information or a daigram.

 Oct 19, 2015
 #2
avatar+118609 
+10
Best Answer

Hi Manas, it is nice to hve your on the forum  laugh

This is a great question :))     

 

the super spy james bond is riding on his super bike at a uniform speed of 60 kmph towards a rocket launcher in chinese territory.the road is straight and horizontal.when he reaches the point x on his way he look at the top of the rocket launcher and finds the angle of elevation to be 30 degree. after 1min. and 30 sec.he looks at the top of the rocket launcher again and finds the angle of elavation is now increased by 30 degree. he takes out a powerful bomb from his pocket and thinks how many sec. he has now,to reach the base of the rocket launcher

can you help him and tell him how many sec. he has?

 

He is travelling at  60mph

This converts to  1000m/s

 

When he starts thinking, which is instantaneous becsues he IS James Bond,  let him be y metres from his target.

Ther the height of the launcher be h

 

In 1 minute and 30 seconds he will have travelled  1.5*1000=1500 metres

so at point X he will be  1500+y  metres frpom the base of the launcer.

 

so

\(tan30=\frac{h}{y+1500} \qquad and \qquad tan60=\frac{h}{y}\\ \frac{1}{\sqrt3}=\frac{h}{y+1500} \qquad and \qquad \frac{\sqrt3}{1}=\frac{h}{y}\\ \frac{y+1500}{\sqrt3}=h \qquad and \qquad \sqrt3\;y=h\\ \frac{y+1500}{\sqrt3} =\sqrt3\;y\\ y+1500 =3\;y\\ 1500 =2\;y\\ y=750 metres\\ \)

 

\(750m\times\frac{1min}{1000m}=0.75 minutes = 45seconds\)

 

 

If conditions stay perfect he has 45 seconds to go.

 

If you have a problem understanding then let me know but make sure you draw the pic and try to work it out first.  :)

Melody Oct 20, 2015
 #3
avatar+44 
+10

melody nice solution but try 30-60-90 theorm for easy steps

 Oct 20, 2015

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