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Question: A body is projected up with velocity \(u\). It reaches the same point in its path at \({t_1}\) and \...

A body is projected up with velocity uu. It reaches the same point in its path at t1{t_1} and t2{t_2} from the time of projection. Then t1+t2{t_1} + {t_2} is:
A. 2ug\dfrac{{2u}}{g}
B. ug\dfrac{u}{g}
C. 2ug\sqrt {\dfrac{{2u}}{g}}
D. ug\sqrt {\dfrac{u}{g}}

Explanation

Solution

Here we have to apply the second equation of motion to get the answer.
The second motion equation gives time to the distance travelled by a body.
Then to get the result, we have to solve the second motion equation.

Complete step by step answer:
The second equation of motion is given by:
s=ut+12at2s = ut + \dfrac{1}{2}a{t^2}
Where
s=s = distance travelled
u=u = initial velocity
t=t = time taken
a=a = acceleration
Here s=hs = h , height of the body
t=t1+t2t = {t_1} + {t_2}

So, the equation for the projection becomes:
s=ut+12gt2 12gt2ut+h=0 12g(t1+t2)2ut+h=0  s = ut + \dfrac{1} {2}g{t^2} \\\ \dfrac{1} {2}g{t^2} - ut + h = 0 \\\ \dfrac{1} {2}g{\left( {{t_1} + {t_2}} \right)^2} - ut + h = 0 \\\
On solving the values for t1{t_1} and t2{t_2} , we get:
Sum of the roots
=ba =ug2 =2ug  = \dfrac{{ - b}} {a} \\\ = \dfrac{u} {{\dfrac{g} {2}}} \\\ = \dfrac{{2u}} {g} \\\
Hence, t1+t2=2ug{t_1} + {t_2} = \dfrac{{2u}}{g}

So, the correct answer is “Option A”.

Additional Information:
- Projectile is a body tossed into the vertical plane with an initial velocity and then travels in two dimensions without being driven by any piston or gasoline under the motion of gravity alone.
- Projectile movement is a two-dimensional occurrence of movement. Any two-dimensional movement case can be tackled one along the x-axis and the other along the y-axis, into two instances of one-dimensional movement. The two instances can be analysed as two instances of one-dimensional motion. The results of two instances can be compared to see the net effect using vector algebra. Vertical and horizontal developments are free of one another.
- The horizontal distance travelled by the body's projectile motion is considered the range of the projectile.

Note:
Here we have to pay attention while finding the roots of the equation. For that we have to remember the rules of quadratic equations. The amount of the quadratic condition, separated by the main coefficient, is equivalent to the nullification of the second term's coefficient. Its acceleration is considered the acceleration of projectiles. A Projectile’s course is called its trajectory.