Chapter 2, Problem 44P

To determine

The height above the window from which the stone falls

Answer to Problem 44P

2.1 m.

Explanation of Solution

Given:

Initial velocity of the stone above the top of window where it was dropped from $=v_o=0\frac{\text{m}}{\text{s}}$

Height above the top of window from where the stone is dropped $=H$

Vertical displacement of the stone before reaching the top of window $=\Delta y=-H$

Initial velocity of the stone at the top of window $=v_i$

Height of the window $=h=2.2\text{m}$

Initial position of the stone at the top of the window $=y_i=h=2.2\text{m}$

Final position of the stone at the bottom of the window $=y_f=0\text{m}$

Acceleration experienced by the stone during free fall $=a=g=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time taken by the stone to pass the window $=t=0.28\sec$

Formula Used:

Position of an object at any time during constant accelerated motion is given as

  $y_f=y_i+v_{i}t+(0.5)a{t}^2$

Displacement during accelerated motion of an object is given as

  $\Delta y=\frac{v_f^2-v_o^2}{2a}$

Calculation:

Consider the motion of the stone in free fall while passing by the window.

Substituting the values,

  $\begin{array}{l}0=2.2+v_i(0.28)+(0.5)(-9.8){(}^{0.28}\\ v_i=-6.4\frac{\text{m}}{\text{s}}\end{array}$

Consider the motion of the stone before it reach the top of window.

Displacement of the stone at the top of window relative to initial starting point is given as

  $\Delta y=\frac{v_f^2-v_o^2}{2a}$

Substituting the values

  $\begin{array}{l}-H=\frac{{(-6.4)}^2-{(0)}^2}{2(-9.8)}\\ H=2.1\text{m}\end{array}$

Conclusion:

Therefore, the height above the window from which the stone falls is 2.1 m.