Question

A body A is thrown up vertically from the ground with a velocity $v0$ and another body B is simultaneously dropped from a height H. They meet at a height $\dfrac{H}{2}$ if $v0$ is equal to
$\left( A \right)\sqrt {2gH}$
$\left( B \right)\sqrt {gH}$
$\left( C \right)\dfrac{1}{2}gH$
$\left( D \right)\dfrac{{2g}}{H}$

Answer 1

First make a rough diagram. Then use the displacement formula from the rectilinear motion equation to find the distance traveled by the two bodies at a height of $\dfrac{H}{2}$. If both of them meet at a height $\dfrac{H}{2}$ we can equate the value which we get from the displacement formula.Then we get the value of time taken using that time on further solving we can find the value of $v0$.

Complete step by step solution:
As per the given problem body A is thrown up vertically from the ground with a velocity $v0$ and another body B is simultaneously dropped from a height H hence the initial velocity of body B is equal to zero. They meet at a height $\dfrac{H}{2}$ .
We need to calculate the initial value of the body A that is $v0$so that both the bodies meet at height $\dfrac{H}{2}$.
Let us take two cases, one for body A and the other for body B.
Case I:
Using rectilinear displacement formula we will get,
$s = ut + \dfrac{1}{2}a{t^2}$
Where,
Distance travelled by the body is equal to s.
Time taken is equal to t.
Acceleration of the body is equal to a.
Initial velocity of the body is equal to u.
Now from the problem we know,
$s = \dfrac{H}{2}$,
$u = v0$
As the body is moving against the gravity, acceleration due to gravity is taken as negative.
$a = - g$
Now putting this values in the rectilinear formula we will get,
$\dfrac{H}{2} = v0t + \dfrac{1}{2}\left( { - g} \right){t^2}$
$\Rightarrow \dfrac{H}{2} = v0t - \dfrac{1}{2}g{t^2} \ldots \ldots \left( 1 \right)$
Case 2:
Using rectilinear displacement formula we will get,
$s = ut + \dfrac{1}{2}a{t^2}$
Where,
Distance travelled by the body is equal to s.
Time taken is equal to t.
Acceleration of the body is equal to a.
Initial velocity of the body is equal to u.
Now from the problem we know,
$s = \dfrac{H}{2}$,
$u = 0$
As the body is moving along the gravity, acceleration due to gravity is taken as positive.
$a = g$
Now putting this values in the rectilinear formula we will get,
$\dfrac{H}{2} = \left( 0 \right)t + \dfrac{1}{2}\left( g \right){t^2}$
$\Rightarrow \dfrac{H}{2} = \dfrac{1}{2}g{t^2} \ldots \ldots \left( 2 \right)$
AS per the problem they meet at a height $\dfrac{H}{2}$.
Hence equating equation $\left( 1 \right)$ and $\left( 2 \right)$ we will get,
$v0t - \dfrac{1}{2}g{t^2} = \dfrac{1}{2}g{t^2}$
Taking the same terms to one side we will get,
$v0t = \dfrac{1}{2}g{t^2} + \dfrac{1}{2}g{t^2}$
$v0t = g{t^2}$
Cancelling the common terms we will get,
$v0 = gt$
Rearranging the equation we will get,
$t = \dfrac{{v0}}{g}$
Putting the value of time in equation $\left( 2 \right)$ we will get,
$\dfrac{H}{2} = \dfrac{1}{2}g{t^2}$
$\Rightarrow \dfrac{H}{2} = \dfrac{1}{2}g{\left( {\dfrac{{v0}}{g}} \right)^2}$
On further solving we will get,
$\dfrac{H}{2} = \dfrac{1}{2}g\dfrac{{v{0^2}}}{{{g^2}}}$
Canceling the common terms we will get,
$H = \dfrac{{v{0^2}}}{g}$
Rearranging the terms we will get,
$v{0^2} = gH$
$\Rightarrow v0 = \sqrt {gH}$
Therefore the correction option is $\left( B \right)$.

Note:
Be careful after finding it further we have to solve this problem because as per the option it is given that the initial velocity of body A should be in relation with acceleration due to gravity and height. Also note that we can put the value of t in any of the equation, here in the solution we used it on equation $\left( 2 \right)$ but you can also put this value in the equation $\left( 1 \right)$ to find the initial velocity of the body A.