Question

A particle is projected vertically upwards from the ground with an initial velocity u.
a.Find the maximum height, H, the particle will attain and time T that it will take to return to the ground.
b.What is the velocity when the particle returns to the ground?
c. What is the displacement and distance travelled by the particle during this time of the whole motion?

Answer 1

a. Consider the motion from A to B:

s = +H (final point lies above the initial point), initial velocity = u, final velocity v = 0.

Let the time taken to go from A to be $t_1$

Using v = u - gt, we get

0 = u - g$t_1$ $\Rightarrow$ $t_1=\frac{u}{g}$

Using s = ut - (1/2)g$t^2$,

H = u$\left(\frac{u}{g}\right)$ - $\frac{1}{2}g$ ${\left(\frac{u}{g}\right)}^2$ = $\frac{u^2}{2g}$

So the maximum height attained is H = $\frac{u^2}{2g}$

Consider the return motion from B to A:

s = -H (final point lies below the initial point)

u = 0 (at point B, velocity is zero)

Let time taken to go from B to A be $t_2$. We have

$t_2$ = $\sqrt{\frac{2H}{g}}$ =$\sqrt{\frac{2u^2}{g2g}}$ = $\frac{u}{g}$

Here $t_1$ is known as the time of ascent and $t_2$ is known as the time of descent. We can see that

Time of ascent = Time of descent = $\frac{u}{g}$

Total time of flight, T = $t_1$ + $t_2$ =$\frac{u}{g}$

Here time of flight is the time for which the particle remains in the air.

$Alternativemethodtofindthetimeofflight:$

a. Consider the motion from A to B:

s = 0 (initial and final points are same)

Initial velocity = u, time taken =T

Using s = ut - (1/2)$g{t}^2$, we have 0 =uT - (1/2)g$T^2$

$\Rightarrow$ T = $\frac{2u}{g}$

b. Magnitude of velocity on returning the ground Will be same as that of initial velocity but direction will be opposite.

Proof: Let v be the velocity on reaching the ground. Then from previous formulae, we get

$v=-\sqrt{2gH}=-\sqrt{2g\frac{u^2}{2g}}$ $\Rightarrow$ v = - u, hence proved.

c. Displacement = 0, distance travelled = 2H = $\frac{u^2}{2g}$