A body is projected up with velocity $u$. It reaches the same point in its path at ${t_1}$ and ${t_2}$ from the time of projection. Then ${t_1} + {t_2}$ is:
A. $\dfrac{{2u}}{g}$
B. $\dfrac{u}{g}$
C. $\sqrt {\dfrac{{2u}}{g}} $
D. $\sqrt {\dfrac{u}{g}} $
Answer
638.1k+ views
Hint: 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:
$s = ut + \dfrac{1}{2}a{t^2}$
Where
$s = $ distance travelled
$u = $ initial velocity
$t = $ time taken
$a = $ acceleration
Here $s = h$ , height of the body
$t = {t_1} + {t_2}$
So, the equation for the projection becomes:
$
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 ${t_1}$ and ${t_2}$ , we get:
Sum of the roots
$
= \dfrac{{ - b}}
{a} \\
= \dfrac{u}
{{\dfrac{g}
{2}}} \\
= \dfrac{{2u}}
{g} \\
$
Hence, ${t_1} + {t_2} = \dfrac{{2u}}{g}$
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.
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:
$s = ut + \dfrac{1}{2}a{t^2}$
Where
$s = $ distance travelled
$u = $ initial velocity
$t = $ time taken
$a = $ acceleration
Here $s = h$ , height of the body
$t = {t_1} + {t_2}$
So, the equation for the projection becomes:
$
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 ${t_1}$ and ${t_2}$ , we get:
Sum of the roots
$
= \dfrac{{ - b}}
{a} \\
= \dfrac{u}
{{\dfrac{g}
{2}}} \\
= \dfrac{{2u}}
{g} \\
$
Hence, ${t_1} + {t_2} = \dfrac{{2u}}{g}$
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.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

