Show that there are two values of time for a projectile when it i…

Question

Show that there are two values of time for a projectile when it is at same height. Also show that the sum of these two times is equal to the time of flight.OR
Prove the following relations by calculus method:

$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
$\text{v}^2-\text{u}^2=2\text{as}$

✓

Answer

For projectile motion equation for y-coordinates $\text{y}=\text{u}\sin(\theta)\text{t}-\frac{1}{2}\text{gt}^2$ Solving this for t (using quadratic formula) $\text{t}=\frac{\text{u}\sin\theta\pm\sqrt{\text{u}^2\sin^2\theta-2\text{gy}}}{\text{g}}$ Without loss of generalityl, let $\text{t}_1=\frac{\text{u}\sin\theta}{\text{g}}+\frac{\sqrt{\text{u}^2\sin^2\theta-2\text{gy}}}{\text{g}}$ $\text{t}_2=\frac{\text{u}\sin\theta}{\text{g}}-\frac{\sqrt{\text{u}^2\sin^2\theta-2\text{gy}}}{\text{g}}$ for $\text{t}_1+\text{t}_2=\frac{2\text{u}\sin\theta}{\text{g}}$ which is equation of time of flight.OR

Consider an object moving in a straight line with uniform acceleration 'a', let at any instant of time 't', dx be the displacement of the objects.

$\therefore$ instantaneous velocity, $\text{v}=\frac{\text{dx}}{\text{dt}}$ or dx = vdt.
dx = (u + at)dt $(\because$ v = u + at)
let $x_0$ and x be the displacements of the obhect at time 'zero' and 't'.
$\int\limits^{\text{x}}_{\text{x}_0}\text{dx}=\int\limits^{\text{t}}_0(\text{u}+\text{at})\text{dt}$
$=\text{u}\int\limits^{\text{t}}_0\text{dt}+\text{a}\int\limits^{\text{t}}_0\text{t dt}$
$(\text{x})^{\text{x}}_{\text{x}_0}=\text{u}(\text{t})^\text{t}_0+\text{a}\Big(\frac{\text{t}^2}{2}\Big)^\text{t}_0$
$\text{x}-\text{x}_0=\text{ut}+\frac{1}{2}\text{at}^2$
If $x = x_0 = s$ (distance) covered by the object.
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

Consider a particle moving in a straight line with initial velocity 'v' and acceleration 'a'.

Then, $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{dv}}{\text{dx}}\times\frac{\text{dx}}{\text{dt}}=\frac{\text{dv}}{\text{dx}}\times\text{v}$
$\text{a dx}=\text{v dv}$
Integrating,
$\int\limits^{\text{x}}_{\text{x}_0}\text{a dx}=\int\limits^{\text{V}}_\text{u}\text{v dv}$
$\Rightarrow\ \text{a}|\text{x}|^{\text{x}}_{\text{x}_0}=\Big[\frac{\text{v}^2}{2}\Big]^\text{V}_\text{u}$
$\text{a}(\text{x}-\text{x}_0)=\frac{\text{v}^2}{2}-\frac{\text{u}^2}{2}$
$\text{v}^2-\text{u}^2=2\text{a}(\text{x}-\text{x}_0)$
$\text{v}^2-\text{u}^2=2\text{a}\text{s.}$