Question

A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height of 4h. The two stones cross each other after time.
$\begin{aligned} & a)\sqrt{\dfrac{h}{2g}} \\\ & b)\sqrt{\dfrac{h}{8g}} \\\ & c)\sqrt{\dfrac{h}{3g}} \\\ & d)none \\\ \end{aligned}$

Answer 1

Let us find the initial velocity of the stone dropped from the ground. After that, assume the distance travelled by stone thrown from above and thrown from below accordingly and their sum will be equal to the total height. Next, we can write the time in terms of the total height.

Formula used:
$\begin{aligned} & {{v}^{2}}-{{u}^{2}}=2as \\\ & s=ut+\dfrac{1}{2}a{{t}^{2}} \\\ \end{aligned}$

Complete step by step answer:
Let us assume the velocity with which the stone was thrown form below be $u$,
We know, at maximum height, the velocity is equal to zero. Therefore, lets calculate the velocity in terms of height,
$\begin{aligned} & {{v}^{2}}-{{u}^{2}}=2as \\\ & \Rightarrow {{0}^{2}}-{{u}^{2}}=2(-g)\times 4h \\\ & \Rightarrow {{u}^{2}}=8gh \\\ & \Rightarrow u=\sqrt{8gh} \\\ \end{aligned}$
Now, let us assume the distance travelled by stone thrown from above and below be ${{h}_{1}},{{h}_{2}}$respectively.
Now, we can write,
$\begin{aligned} & {{h}_{1}}=0+\dfrac{1}{2}(g){{t}^{2}} \\\ & \Rightarrow {{h}_{1}}=\dfrac{1}{2}g{{t}^{2}} \\\ & \Rightarrow {{h}_{2}}=ut+\dfrac{1}{2}a{{t}^{2}} \\\ & \Rightarrow {{h}_{2}}=\sqrt{8gh}t+\dfrac{1}{2}(-g){{t}^{2}} \\\ & \\\ \end{aligned}$1
Now, we can easily say that the sum of both the heights must be equal to total height.
$\begin{aligned} & {{h}_{1}}+{{h}_{2}}=h \\\ & \Rightarrow \dfrac{1}{2}g{{t}^{2}}+\sqrt{8gh}t+\dfrac{1}{2}(-g){{t}^{2}}=h \\\ & \Rightarrow h=\sqrt{8gh}t \\\ & \Rightarrow t=\sqrt{\dfrac{h}{8g}} \\\ \end{aligned}$

So, the correct answer is “Option A”.

Additional Information: Newton's equation of motion are the mathematical equations that describe position, velocity, or acceleration of the given body relative to a given frame of reference. Newton's second law states that force acting on a body is equal to mass multiplied by the acceleration of the body is the basic equation of motion in classical mechanics. If the force is a function of time, the displacement and velocity can also be represented as a function of time by solving them through a process known as integration. For example, a falling body accelerates at a constant rate acceleration is the rate of change of velocity with respect to time velocity is the time rate change of position. If the force acting on a body is specified as a function of position or velocity, the integration of newton's equation may be more difficult. We can also use differentiation to find the equation of velocity time and acceleration time.

Note: In the above question, in the second case, the height of the stone thrown from the below is negative and height travelled by the stone thrown from the above is taken as positive because the line of reference chose is the ground and the stone is moving towards the ground, i.e., decreasing. Therefore, appropriate signs must be taken.