Question

A body thrown up with a velocity of $98 {m}/{s}$ reaches a point P in its path 7 second after projection. Since its projection it comes back to the same position after:
A.13s
B.14s
C.6s
D.22s

Answer 1

This question can be solved using the Kinematics equations. Firstly, use the Kinematics equation showing relation between velocity and position, to find the maximum height at which the velocity is zero. Then, use the Kinematics equation showing relation between time and position. Using this relation calculate the time taken by the body to reach the maximum height. Double this obtained value to get the time taken by the object to return back from the maximum height. Then, subtract the time to reach point P from the obtained time and get the time taken by the body to come back to the same position since its projection.

Complete Answer:
Given: Initial velocity (u) = $98 {m}/{s}$
Time (${t}_{1}$)= 7s
a= $9.8 {m}/{s}$
At maximum height, ${v}_{1}$=0.
Maximum height can be calculated using the Kinematic equation given as,
${v}^{2}= {u}^{2} +2as$
As the velocity is zero at maximum height, the equation turns to,
${u}^{2}= 2as$
$\therefore s= \dfrac {{u}^{2}}{2a}$ …(1)
Now, time for the body to reach the maximum height can be calculated using the formula,
$s= ut + \dfrac {1}{2}a{t}^{2}$
$s= \dfrac {1}{2}a{t}^{2}$
Substituting equation. (1) in above equation we get,
$\dfrac {{u}^{2}}{2a} = \dfrac {1}{2}a{t}^{2}$
Substituting values in above equation we get,
$\dfrac {{98}^{2}}{2 \times (9.8)} = \dfrac {1}{2}\times (9.8) \times{t}^{2}$
$\Rightarrow 490=4.9{t}^{2}$
Rearranging the above equation we get,
${t}^{2} = \dfrac {490}{4.9}$
$\Rightarrow {t}^{2} =100$
$\Rightarrow t= 10s$
Total time required for the body to comeback from the maximum height is given by,
$t=2 \times 10$
$\Rightarrow t=20s$
Now, the time taken by the body to reach point P from ground and to reach ground from point P is 7s.
Therefore, time taken by the body to come back to the same position P since its projection is given by,
$T= t – {t}_{1}$
Substituting the values we get,
$T= 20 – 7$
$\therefore T= 13s$

Hence, the correct answer is option A i.e. 13s.

Note:
Students should remember the Kinematic equations to solve such kinds of problems. Not only remember the equations but should also know the proper application of each equation. If you carefully select the equation which describes the situation in the problem, then solving the problem will become easier for the students. If you want to find the position at a given time, then you can use the equation, $s= ut + \dfrac {1}{2}a{t}^{2}$. To find the time at a given velocity, use the equation $v=u+at$. To find the position at a given velocity, use the relation ${v}^{2}= {u}^{2} +2as$.