Question

During the last second of its flight, a ball thrown vertically upwards covers one half of the distance covered during the whole flight. The point of the project and the point of landing may or may not be in the same horizontal level. What maximum possible duration of the flight can be obtained? neglect air resistance and assume acceleration of free fall to $10m/{{s}^{2}}$

Answer 1

The solution to this problem is obtained by utilizing equation of motion and also by the speed and direction relationship .Maximum speed of flight depends upon the distance covered by the ball and also the time taken by the ball to cover the distance air resistance is a force caused by air is neglected here

Complete answer:
Maximum time of flight is possible when the distance covered by the ball in the last one second is maximum.
Let us consider the duration of the flight be $T$
The distance travelled by the ball after $t$ second be${{s}_{t}}$
Initial speed of the ball is $u$
So we are given
${{s}_{T}}-{{s}_{T-1}}=\dfrac{{{s}_{T}}}{2}$
${{s}_{T}}=2\times {{s}_{T-1}}$ $\cdots \cdots (1)$
Equation of motion is given as follows
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
${{v}^{2}}={{u}^{2}}+2as$
Maximum height(h) is reached by the ball when $v=0$
$0={{u}^{2}}-2gh$
$h=\dfrac{{{u}^{2}}}{2g}$
The time taken to reach height h is $t=\dfrac{u}{g}$
${{s}_{T}}=h+\dfrac{1}{2}g{{(T-t)}^{2}}$
${{s}_{T-1}}=h+\dfrac{1}{2}g{{(T-t-1)}^{2}}$
Substitute ${{s}_{T}}$ and ${{s}_{T-1}}$ in equation$(1)$
$h+\dfrac{1}{2}g{{(T-t)}^{2}}=2\times (h+\dfrac{1}{2}g{{(T-t-1)}^{2}})$
$\dfrac{{{u}^{2}}}{2g}+\dfrac{1}{2}g{{(T-t)}^{2}}=2\times (\dfrac{{{u}^{2}}}{2g}+\dfrac{1}{2}g{{(T-t-1)}^{2}})$
${{T}^{2}}-(4+2t)T+(4t+2+{{t}^{2}})=0$
Obtaining the solution of $T$ in terms of $t$
$T=\dfrac{4+2t+\sqrt{8-4{{t}^{2}}}}{2}$ $\cdots \cdots (2)$
In order to maximize $T$ neglect square root term
Differentiate $T$ with respect to $t$
$\dfrac{dT}{dt}=0$
$2-\dfrac{4t}{\sqrt{8-4{{t}^{2}}}}=0$
$32-16{{t}^{2}}=16{{t}^{2}}$
We get the value of $t=1$
Substituting this values in equation$(2)$
$T(t=1)=\dfrac{4+2+\sqrt{8-4}}{2}$
$=4\sec$
The maximum possible duration of the flight is$4\sec$

Note:
Students the time of flight is just the double of the maximum height time the maximum time of flight is also determined solely by the initial velocity in the y direction and the acceleration due to gravity. The point of the project and the point of landing may or may not be in the same horizontal level.