Question

Show that time of flight is $\dfrac{{2u}}{g}$.

Answer 1

Hint: We can use the second equation of motion which gives us the relation between time, initial speed, and gravitational force. Arrange the equation and find time of flight in terms of initial speed and gravitational constant.

Complete step by step answer:
For an object which has been projected vertically upwards with some initial velocity let it be $u\dfrac{m}{s}$ , then equation for vertical displacement is given by,
$y = ut - \dfrac{1}{2}g{t^2}$
Where $u$ is the initial velocity, $t$ is the time for which the object is in the air, and $g$ is a gravitational constant. When the distance travelled is zero then, \[y = 0\].
$ \Rightarrow ut - \dfrac{1}{2}g{t^2} = 0$
$ \Rightarrow 2ut - g{t^2} = 0$
\[ \Rightarrow t\left( {2u{\text{ }} - {\text{ }}g} \right) = 0\]
\[ \Rightarrow t = 0,{\text{ }}\dfrac{{2u}}{g}\]
Hence, \[t = \dfrac{{2u}}{g}\] is called the time of flight, this tells us how much time an object has been in the air. The measurement of the time it takes an object, particle, or wave (whether acoustic, electromagnetic, or other) to travel a distance across a medium is known as time of flight . This data can then be used to calculate velocity or path length, or to learn more about the properties of the particle or medium.

Note: We construct three conventional equations of motion, often known as the laws of constant acceleration, in the case of motion with uniform or constant acceleration (one with equal change in velocity in equal intervals of time). These equations govern the motion of a particle by containing the values displacement( $s$ ), velocity (initial and final), time( $t$ ), and acceleration( $a$ ). Only when a body's acceleration is constant and motion is on a straight line can these equations be used.