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.

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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.

Physics
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what does it mean "from start" ?!
from initial point
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.

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Already did that, the problem is, how to merge t1, t2. And the equation of projectile is \[x \tan \alpha - gx ^{2}/2u ^{2}\cos ^{2}\alpha\]
thats the y component
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.
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.
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.
*dropped the minus sign on the g terms.
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!
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.
You are a genius and a lifesaver, just cannot thank you enough! It worked :)

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