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.

Question

y2o2 · Answer

what does it mean "from start" ?!

anonymous · Answer

from initial point

jamesj · Answer

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.

anonymous · Answer

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

anonymous · Answer

thats the y component

jamesj · Answer

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.

jamesj · Answer

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.

jamesj · Answer

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.

jamesj · Answer

*dropped the minus sign on the g terms.

anonymous · Answer

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!

jamesj · Answer

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.

anonymous · Answer

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