Question

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other.
(a) If the faster stone takes $10s$ to return to the ground, how long will it take the slower stone to return?
(b) If the slower stone reaches a maximum height of $H$ , how high (in terms of $H$ ) will the faster stone go?
Assume free fall.

Answer 1

In order to find the solution to the question above, we will be using the concepts and formulas for the motion of a system in one dimension. We will also be using the basic laws of kinematics.

Complete step by step answer:
As mentioned in the question, two stones are thrown at an initial velocity of $u$ and $3u$ in the vertically upward direction.
(a) If the faster stone takes $10s$ to reach the ground, how much time does the slower stone take to reach the ground. The displacement of the stones will be zero as they are thrown from the ground and return at the same point.
For faster stone, we can write the equation as,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
We know that the time taken by the fast stone is $10s$ . The gravitational direction is in the opposite direction of motion and is $9.8\dfrac{m}{{{s}^{2}}}$
$\begin{aligned} & \Rightarrow 0=3ut-\dfrac{1}{2}g{{t}^{2}} \\\ & \Rightarrow 3(10)u=\dfrac{1}{2}(9.8){{(10)}^{2}} \\\ & \Rightarrow u=\dfrac{1}{30}(490) \\\ & \Rightarrow u=16.33\dfrac{m}{s} \\\ \end{aligned}$
Using this, the time taken by the slower stone is calculated as
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
$\begin{aligned} & \Rightarrow 0=(16.33)t-\dfrac{1}{2}(9.8){{t}^{2}} \\\ & \Rightarrow (16.33)t=\dfrac{1}{2}(9.8){{t}^{2}} \\\ & \Rightarrow t=\dfrac{16.33(2)}{9.8} \\\ & \Rightarrow t=3.33s \\\ \end{aligned}$
The time of flight for the slower stone is $3.33s$
(b) The velocity of the stone when it reaches the maximum height becomes zero and hence the equation for maximum height can be given by
$\begin{aligned} & 2as={{v}^{2}}-{{u}^{2}} \\\ & \Rightarrow -2gh=0-{{u}^{2}} \\\ & \Rightarrow h=\dfrac{{{u}^{2}}}{2g} \\\ \end{aligned}$
For, slower stone, the maximum height reached is $H$ and initial velocity is u. Therefore,
$\Rightarrow H=\dfrac{{{u}^{2}}}{2g}$
Similarly, for faster stone with initial velocity $3u$ , the maximum height reached is given by
$\begin{aligned} & \Rightarrow h=\dfrac{{{(3u)}^{2}}}{2g} \\\ & \Rightarrow h=\dfrac{9{{u}^{2}}}{2g} \\\ & \Rightarrow h=9H \\\ \end{aligned}$
As $\left[ H=\dfrac{{{u}^{2}}}{2g} \right]$
We can conclude from the equation that the maximum height reached by the faster stone is nine times the height of slower stone.

Additional Information:
First and foremost, this must be understood by us in context. Free fall, according to Newton, is motion that occurs in the absence of external forces other than gravity.
Gravity is not a force, as Einstein demonstrated. It is due to the curvature of space-time. Free fall is motion along a geodesic, which is the four-dimensional counterpart of a straight line, in terms of relativity.
Anything moving in free space can be defined as free fall. It may be something that is dropping as a result of gravity. Anything in orbit is the most common form of free fall. The ISS, like the Moon, is in free fall. You would be weightless if you are in a ship in free fall.
In a gravitational field, most objects in free fall follow a geodesic. As a result, gravity is to blame for the majority of free fall situations. Gravity is the only force acting on an entity in free fall. There is no air resistance in free fall. In free fall, regardless of mass, all objects fall at the same rate of acceleration.

Note:
It is important to keep in mind the direction of motion and the direction of acceleration due to gravity. As in the above question, the direction of motion and the gravitational acceleration are in opposite directions to each other, hence the negative sign.