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 $$20m{s}^{-1}$$. 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

Hint: It is given that a stone is initially dropped from a tower of height s and later another stone is dropped from the same height with a velocity of $$20m{s}^{-1}$$. So , the second stone is thrown at an initial velocity and will go down experiencing additional velocity along with gravitational use. Using equations of motion, find the point where stone 2 will overtake the first one.
Formula used:
$$s=ut+{\displaystyle \frac{1}{2}}g{t}^2$$

Complete step by step answer :
Look at the given data. We know that there is a stone of mass m initially dropped from a height.At given time t, the stone travels a distance $$s$$.Since the stone is said to be dropped from rest position, the initial velocity will be zero. Now the equation shapes up as,
$$s=(0)t+{\displaystyle \frac{1}{2}}(9.8)t^2$$-----(1)
Now, the second stone is dropped with an initial velocity of $$20m{s}^{-1}$$ from the same height $$s$$a second later. Since the initial velocity of the stone is higher than that of the first stone, it is assumed that it travels faster than the first stone. Now, the distance travelled by the second stone in time (t-1) since t is the time taken by the first stone, is given as,
$$\Rightarrow s=(20)(t-1)+{\displaystyle \frac{1}{2}}(9.8)(t-1{)}^2$$
Now, at the instance of overtaking, let us consider that the distances are the same. Now we can equate equation (1) and (2) and find the time of overtake.
$$\Rightarrow {\displaystyle \frac{1}{2}}g{t}^2=(20)(t-1)+{\displaystyle \frac{1}{2}}(9.8)(t-1{)}^2$$
Cancelling out common terms and rearranging the equation we get,
$$\Rightarrow {\displaystyle \frac{1}{2}}g{t}^2=(20t-20)+{\displaystyle \frac{1}{2}}g(t^2-2t+1)$$
On further simplification ,we get,
$$\Rightarrow {\displaystyle \frac{1}{2}}g{t}^2=(20t-20)+{\displaystyle \frac{1}{2}}g{t}^2-gt+{\displaystyle \frac{1}{2}}g$$
Cancelling out the common term, we get,
$$\Rightarrow 0=20t-20-gt+{\displaystyle \frac{1}{2}}g$$
Substituting g as $$10m{s}^{-2}$$, we get,
$$\Rightarrow 0=20t-20-10t+5$$
Taking like terms on same side we get,
$$\Rightarrow 15=10t$$
$$\Rightarrow t=1.5s$$
Now, substitute this value to find the point of overtaking. Substitute t in equation (1) we get
 $$\Rightarrow s={\displaystyle \frac{1}{2}}(10)(1.5{)}^2$$
$$\Rightarrow s=11.25m$$

Hence, Option(c) is the right answer

Note: We can solve this also by finding the distance travelled by stone 1 in one second and assume the value of the distance travelled by stone 2 as distance of stone 1 plus some variable. Using second equations of motion, equate and solve for x.