Q48P - Fundamentals Of Physics

Question

A hoodlum throws a stone vertically downward with an initial speed of $\mathbf{12}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$ from roof of a building, 30 m above the ground. (a) How long does it take the stone to reach the ground? (b) What is the speed of stone at impact?

Step-by-Step Solution

Answer

(a) Time taken by stone to reach the ground is 1.54 sec.

(b) Velocity of the stone when it reaches the ground is $27.1 \mathrm{m}/\mathrm{s}$.

Given information

The given information is,

$\text{InitialSpeed}\left(v_0\right)=12.0\text{m}/\text{s}$

$\text{HeightofBuilding}\left(s\right)=30.0\text{m}$

Understanding the concept

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Also, it is based on the concept of free fall which is the motion of a body where gravity is the force acting upon it. The stone has been given the vertically downward velocity. The acceleration due to gravity would be acting in the same direction. Using the kinematic equation, find the time it would take to cover the height of the building. Also, using another kinematic equation, find the speed of the stone to reach the ground.

 

Formula:

The velocity in kinematic equations can be expressed as,

 

${v_f}^2={v_0}^2+2as$

$v_f=v_0+at$

$\text{Acceleration}=\text{GravitaionalAcceleration}=9.8\text{m}/{\text{s}}^2$

Determination of speed of stone hits ground

Consider downward as positive and upward as negative.

Calculate velocity (Part b), which can then be used in the calculation of time. 

Using Kinematic Equation

$\begin{array}{rcl}{v_f}^2& =& {v_0}^2+2as\\ {v_f}^2& =& 12.0^2 {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}+2\times 9.8 {\text{m/s}}^{\text{2}}\times 30.0 \text{m}\\ {v_f}^2& =& 732 m^2/s^2\\ v_f& =& 27.05\\ & \approx & 27.1 \text{m}/\text{s}\end{array}$

Thus, the stone would be moving with $27.1 \mathrm{m}/\mathrm{s}$ velocity, when it is about to hit the ground.

Determination of time taken by stone to hits the ground

Using Kinematic Equation

$\begin{array}{rcl}{v}_f& =& v_0+at\\ 27.1 \text{m/s}& =& 12.0 \text{m/s}+9.8 {\text{m/s}}^{\text{2}}\times t\\ t& =& \frac{15.1 \text{m/s}}{9.8 {\text{m/s}}^{\text{2}}}\\ t& =& 1.54\text{sec}\end{array}$

The stone would take $1.54 s$ to reach the ground.