Q: m1 and m2 are thrown separately with an angle of [tex]\theta1[/tex] and [tex]\theta2[/tex] respectively from the same place from the ground. They both have the same initial velocity, V. How long should the time difference be if m1 and m2 are to meet in space?(adsbygoogle = window.adsbygoogle || []).push({});

My work:

For m1:

x1=Vcos[tex]\theta[/tex]1*t1

y1=Vsin[tex]\theta1[/tex]*t1-0.5g[tex]t1^2[/tex]

For m2:

x2=Vcos[tex]\theta2[/tex]*t2

y2=Vsin[tex]\theta2[/tex]*t2-0.5g[tex]t2^2[/tex]

Therefore,

x1=x2

=>Vcos[tex]\theta1[/tex]*t1=Vcos[tex]\theta2[/tex]*t2

=>t1/t2=cos[tex]\theta2[/tex]/cos[tex]\theta1[/tex]*t2

And

y1=y2

=>Vsin[tex]\theta1[/tex]*t1-0.5g[tex]t1^2[/tex]=Vsin[tex]\theta2[/tex]*t2-0.5g[tex]t2^2[/tex]

=>V[sin[tex]\theta1[/tex]*t1-sin[tex]\theta2[/tex]*t2]=0.5g[tex]t1^2[/tex]-[tex]t2^2[/tex]

=>V[(sin[tex]\theta1[/tex]*cos[tex]\theta2[/tex]/cos[tex]\theta1[/tex])1)] =0.5g[cos[tex]\theta2^2[/tex]*[tex]t2^2[/tex]/cos[tex]\theta1[/tex]]

=>2V[(sin[tex]\theta1[/tex]*cos[tex]\theta2[/tex]-cos[tex]\theta1[/tex])/cos[tex]\theta1[/tex])]/g=t2*[cos[tex]\theta2^2[/tex]-cos[tex]\theta1^2[/tex]]/cos[tex]\theta1^2[/tex]]

=> t2=2V*[sin[tex]\theta1[/tex]*cos[tex]\theta2[/tex]]*cos[tex]\theta1/tex]/[g*(cos[tex]\theta2^2[/tex]-cos[tex]\theta1^2[/tex])

so

t1= t2=2V*[sin[tex]\theta1/tex]*cos[tex]\theta2[/tex]]*cos[tex]\theta1/tex]/[g*(cos[tex]\theta2^2[/tex]-cos[tex]\theta1^2[/tex])*cos[tex]\theta[/tex]

but t2-t1 doesn't equal the book's answer:

2vsin[tex]\theta1-\theta2[/ex]/{g(cos[tex]\theta1[/tex]+cos[tex]\theta2/tex]}

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# Homework Help: Time difference in projectile motion

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