## arcticf0x Group Title A particle during projectile motion reaches height h in time t1. Again it reaches this height 'h' in time t2 measured from start. Show that the height of point 'h' is 1/2 gt1t2. 2 years ago 2 years ago

1. y2o2

what does it mean "from start" ?!

2. arcticf0x

from initial point

3. JamesJ

What's the equation of motion of the projectile? Let y be its height above the point from which it's thrown. Then $y(t) = v_0t - \frac{g}{2}t^2$ Now you're told that for two values of time, $$t = t_1, t_2$$ $y(t_1) = y(t_2) = h$ Use these equations to solve for h.

4. arcticf0x

Already did that, the problem is, how to merge t1, t2. And the equation of projectile is $x \tan \alpha - gx ^{2}/2u ^{2}\cos ^{2}\alpha$

5. arcticf0x

thats the y component

6. JamesJ

The equation I've written down is just the vertical component, the y. And that's all you need. The x component isn't necessary.

7. JamesJ

Substitute t = t1 and t2 into that equation and you have two simultaneous equations in the variables v_0 and h. Now solve them for h.

8. JamesJ

I.e., $\frac{g}{2}t_1^2 + v_0 t_1 = h$ $\frac{g}{2}t_2^2 + v_0 t_2 = h$ Solve for h.

9. JamesJ

*dropped the minus sign on the g terms.

10. arcticf0x

yeah i am almost there now. I eliminated v0 and the one equation is left in the terms i need. Its not rearraning easily but i will leave it to that because i have got the idea. Thanks again JamesJ!

11. JamesJ

Multiply the first equation by t_2; the second by t_1. Then subtract the second equation from the first and the v_0 terms will cancel.

12. arcticf0x

You are a genius and a lifesaver, just cannot thank you enough! It worked :)