Question

A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity 20m/s. The second stone will overtake the first after travelling a distance of $(g=10m/{{s}^{2}})$
A) 13m
B) 15m
C) 11.25m
D) 19.5m

Answer 1

Here in the given question there are two stones which are dropped one after another. The first stone is dropped with an initial velocity zero and after one second another is dropped with initial velocity 20m/s. We have to find the distance at which the second stone overtakes the first which can be done by using distance formula in terms of acceleration, velocity and time.

Formula used:
$s=ut+\dfrac{1}{2}g{{t}^{2}}$
Complete answer:
Let us draw a simple diagram for the given question.

Here two stones, 1 and 2, are dropped from the top of the tower which are moving vertically downwards. Therefore the acceleration of both the stone would be due to the gravitational pull and it can be taken as g, which is acceleration due to gravity.
We have to find the distance at which stone 2 overtakes 1. Now at the point where 2 overtakes 1 distance travelled by them will be the same. That is distance travelled by 1 will be equal to 2.
Now the distance travelled by an object having acceleration a and initial velocity u in time t is given as
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
For stone 1, the let say initial velocity is zero and acceleration is given by g as we discussed above. Let say it takes time t to travel cover the distance s, then by using distance formula

& s=(0)t+\dfrac{1}{2}g{{t}^{2}} \\\ & \Rightarrow s=\dfrac{1}{2}g{{t}^{2}} \\\ \end{aligned}$$ Now value of g is given$$10m/{{s}^{2}}$$, substituting this we get $$\begin{aligned} & s=\dfrac{1}{2}\left( 10 \right){{t}^{2}} \\\ & \Rightarrow s=5{{t}^{2}}\text{ }..........\text{(i)} \\\ \end{aligned}$$ For stone 2, the initial velocity is 20m/s as it was thrown by 20m/s and it will accelerate g as mentioned earlier. As it was thrown after one second of stone 1, therefore time taken to travel distance s would be one second less than time taken by stone 1 i.e. (t-1), hence from distance formula we have, $$\begin{aligned} & s=(20)\left( t-1 \right)+\dfrac{1}{2}g{{\left( t-1 \right)}^{2}} \\\ & \Rightarrow s=20t-20+\dfrac{1}{2}g\left( {{t}^{2}}-2t+1 \right) \\\ & \Rightarrow s=20t-20+\dfrac{1}{2}g{{t}^{2}}-gt+\dfrac{1}{2}g \\\ \end{aligned}$$ Substituting given $$g=10m/{{s}^{2}}$$ we get $$\begin{aligned} & \Rightarrow s=20t-20+\dfrac{1}{2}\left( 10 \right){{t}^{2}}-\left( 10 \right)t+\dfrac{1}{2}\left( 10 \right) \\\ & \Rightarrow s=20t-20+5{{t}^{2}}-10t+5 \\\ & \Rightarrow s=10t-15+5{{t}^{2}}\text{ }.........\text{(ii)} \\\ \end{aligned}$$ If s is the distance at which the stone 1 overtakes stone 2, then equation (i) will be equal to equation (ii), hence we can write $$\begin{aligned} & 5{{t}^{2}}=10t-15+5{{t}^{2}} \\\ & \Rightarrow 10t-15+5{{t}^{2}}-5{{t}^{2}}=0 \\\ & \Rightarrow 10t-15=0 \\\ & \Rightarrow 10t=15 \\\ & \Rightarrow t=\dfrac{15}{10} \\\ & \Rightarrow t=1.5s \\\ \end{aligned}$$ Now substituting value of t in equation (i) we get $$\begin{aligned} & s=5{{(1.5)}^{2}} \\\ & \Rightarrow s=5(2.25) \\\ & \Rightarrow s=11.25m \\\ \end{aligned}$$ **Hence option C is correct.** **Note:** Here the time taken by stone 1 and 2 are different and distance same as the velocity at which it was thrown or dropped was different i.e. there initial velocity. Also initial velocity of stone 1 is taken zero as it was not given in the question and if before dropping it, the stone would be at rest and so it will have zero initial velocity.