Kinematics stone thrown upwards past a point, show the following.

Question

I know I should be able to do this, but I have tried for 3 hours and can't do it. I know its simple but it's driving me mad.

A particle is projected vertically upwards with speed $ u_{0}$ and passes through a point that is a distance $ h $ above the point of projection at time $t_{1}$ going up and $t_{2}$ coming down. Show that $g t_{1} t_{2} = 2 h$.

I am assuming the time taken for the stone to go from point $h$ up to the max height is equal to the time taken for the stone to fall from the max height down to the point $h$. This time being $\frac{t_{2}-t_{1}}{2}$. I've used the SUVAT equations ... many times ,and the answer won't deliver. Any help appreciated.

Answer 1

The kinetic and potential energies are given by $$T=\frac 12 m \dot h^2$$ $$V=m\,g\,h$$ By applying Lagrangian mechanics the equation of motion can be found as $$\ddot h=-g\Rightarrow h=-\frac g2 t^2+a\,t+b$$ By initial conditions $\dot h(0)=u_0$ and $h(0)=0$ $$h=-\frac g2 t^2+u_0\,t$$ It can be described as a quadratic equation $$\frac g2 t^2-u_0\,t+h=0$$ which has the following roots $$t_1=\frac {u_0+\sqrt{u_0^2-2\,h\,g}}{g}\qquad t_2=\frac {u_0-\sqrt{u_0^2-2\,h\,g}}{g}$$ and $$t_1 t_2=\frac {u_0+\sqrt{u_0^2-2\,h\,g}}{g}\frac {u_0-\sqrt{u_0^2-2\,h\,g}}{g}=\frac{u_0^2-u_0^2+2\,h\,g}{g^2}$$ $$t_1\,t_2=\frac{2h}g\Rightarrow g\,t_1\,t_2=2h$$

Answer 2

Rewriting the constant-acceleration equation $s=ut+\frac12at^2$ as $$\frac12at^2+ut-s=0$$ and applying Vieta's Formula gives $$t_1t_2=\frac{-s}{\frac12 a}\\-at_1t_2=2s.$$ Define the reference point as the point of projection and the positive direction as upwards so that

$a=-g$ and $s=h$, and we are done: $$gt_1t_2=2h.$$

Answer 3

You only need the one kinematics (SUVAT) equation. $s = vt - \frac12 at^2$. Here $v$ is the final velocity after acceleration $a$ from an initial velocity $u$ for time $t$ over a displacement $s$.

Set $s_1 = s_2 = +h, a = - g$ (upward positive, downward negative). To avoid confusion, use $+w$ to represent the velocity at $t_1$. By conservation of mechanical energy (lack of drag preserves symmetric parabolic motion), we have velocity $-w$ af $t_2$

We now have the equation system:

$h = wt_1 +\frac 12gt_1^2$

$h = -wt_2 +\frac 12gt_2^2$

Subtract the second from the first to give:

$w(t_1 + t_2) + \frac 12g(t_1^2 - t_2^2) =0$

which, after applying the difference of squares identity, becomes:

$w(t_1 + t_2) + \frac 12g(t_1+t_2)(t_1-t_2)=0$

Since $(t_1 + t_2) \ne 0$, we get (after division) $w= -\frac 12g(t_1-t_2)$

Now add the two original equations to get:

$2h = w(t_1-t_2) + \frac 12g(t_1^2+t_2^2)$

Put $w= -\frac 12g(t_1-t_2)$ into that to get:

$2h =-\frac 12g(t_1-t_2)^2 + \frac 12g(t_1^2+t_2^2)$

Expand and simplify to: $gt_1t_2=2h$, as required.