Question

A stone dropped from the top of a tower travels $4.9{\text{ }}m$ in the last second, then the velocity of the stone on reaching the ground is
A) $19.6\,m/s$
B) $9.8\,m/s$
C) $4.9\,m/s$
D) $29.4\,m/s$

Answer 1

Hint : In this solution, we will use the equation of kinematics to determine the velocity of the stone before the last second using the second equation of kinematics and then we will determine the velocity of the stone when it reaches the ground using the first law of kinematics.

Formula used: In this solution, we will use the following formulae:
Second equation of kinematics: $d = ut + \dfrac{1}{2}a{t^2}$ where $d$ is the distance travelled by the stone with initial velocity $u$ under acceleration $a$ in time $t$
First equation of kinematics: $v = u + at$ where $v$ is the final velocity of the stone.

Complete step by step answer
We’ve been given that a stone dropped from the top of a tower travels $4.9{\text{ }}m$ at the last second. Let us determine the distance the stone travels in this last second using the second law of kinematics which tells us
$d = ut + \dfrac{1}{2}a{t^2}$
Substituting $d = 4.9\,m$ and $t = 1$ , $a = g = 9.8\,m/{s^2}$ , we get
$4.9 = u(1) + 4.9{(1)^2}$
$\Rightarrow u = 0\,m/s$
Since we obtained the initial velocity as zero, this means that the stone is dropped from the top of the tower and it falls to the ground in the first second itself. So, in this case, the last second of its drop is the first second itself since it falls to the ground in that time. The velocity when it touches the ground can be calculated as:
$v = u + at$
Substituting the values we get,
$v = 0 + 9.8(1)$
$\therefore v = 9.8\,m/s$ which corresponds to option (B).

Note
Since the stone drops the entire height of the building in one second and the distance travelled by it is $4.9{\text{ }}m$ , that is also the height of the building. Since the stone is free falling in absence of air resistance, we should take the acceleration of the stone as $a = g = 9.8\,m/{s^2}$ .