# Time difference in projectile motion

1. Aug 4, 2006

### asdf1

Q: m1 and m2 are thrown separately with an angle of $$\theta1$$ and $$\theta2$$ 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?

My work:

For m1:
x1=Vcos$$\theta$$1*t1
y1=Vsin$$\theta1$$*t1-0.5g$$t1^2$$

For m2:
x2=Vcos$$\theta2$$*t2
y2=Vsin$$\theta2$$*t2-0.5g$$t2^2$$

Therefore,
x1=x2
=>Vcos$$\theta1$$*t1=Vcos$$\theta2$$*t2
=>t1/t2=cos$$\theta2$$/cos$$\theta1$$*t2

And
y1=y2
=>Vsin$$\theta1$$*t1-0.5g$$t1^2$$=Vsin$$\theta2$$*t2-0.5g$$t2^2$$

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

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

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

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

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

but t2-t1 doesn't equal the book's answer:
2vsin$$\theta1-\theta2[/ex]/{g(cos[tex]\theta1$$+cos$$\theta2/tex]} Last edited by a moderator: Aug 4, 2006 2. Aug 4, 2006 ### Päällikkö The equations you wrote are quite hard to read, I'll try anyways. Certainly t1 = t2 cannot hold or the masses would've been thrown at the same time. Instead of your equations (I'll use O for theta) eg. x1 = vt1cos(O1), try the following: x1 = v(t - t1)cos(O1) I ended up with a huge equation for t2 - t1, but (with the help of Maple ) I managed to reduce it to your book's answer. 3. Aug 9, 2006 ### asdf1 why t-t1? agh... i made a mistake with my typing! it's t1/t2 not t1=t2... sorry about that! 4. Aug 9, 2006 ### Päällikkö t1 here is a constant. It therefore represents the instant (t = t1) the mass 1 is thrown. You are left with only one timevariable t, which is equal in all equations. Another way you could do it, is to set a constant [tex]\Delta t = t_2 - t_1$$
This is pretty much equivalent to the stuff with t - t1

Using two independent time variables, t1 and t2 is first of all physically difficult to comprehend. Also, this way you fail to use the information that the masses must be at the spot (x0, y0) at the same time.

By the way, you've made a mistake with the first t1/t2 equation, there shouldn't be a t2 on the right hand side of the equation.

5. Aug 9, 2006

### nrqed

I cleaned up a bit the equations (suggestions: do not break the eqs into pieces enclosed with tex tags and pieces in normal format, put everything in one single tex quote. Also, use the underscore symbol for indices like 1 and 2)

I don't understand how you got to that last line. I am not sure what the "1" sy,bl by itself means, but even if it is t1 or t1^2, the equation is wrong. Starting from that line the algebra does not seem to make sense.

What you must do is express everywhere in the above equation t1 in terms of t2 (or vice versa) using the equation you got from the x axis. then you get
$$V t_2 ( {sin \theta_1 cos \theta_2 \over cos \theta_1} - sin \theta_2) = {1\over 2} g t_2^2 ({cos^2 \theta_2 \over cos^2 \theta_1}-1)$$
Isolating t2 you should be able to show that you get (after a bit of algebra)
$$t_2 = {2 V \over g} {(sin \theta_1 cos \theta_2 - sin \theta_2 cos \theta_1 ) cos \theta_1 \over cos^2 \theta_2 - cos^2 \theta_1 }$$
Then you can find t1 and subtract the two expressions. Notice that $cos^2 \theta_2 - cos^2 \theta_1 = (cos \theta_2 + cos \theta_1) (cos \theta_2 - cos \theta_1)$ which will prove useful in getting the final answer.

Patrick

Last edited: Aug 9, 2006
6. Aug 14, 2006

### asdf1

ok!
thank you very much! ^_^