Chapter 2, Problem 42P

Solution

a.

The speed of stone.

  $v=10.41$ m/sec

Given:

Initial speed is 18m/sec vertically upwards.

Height is 11 m.

Formula used:

  $v^2=u^2-2gh$ here v is the speed at height h, u is initial speed , h is height , g is acceleration due to gravity.

Calculation:

Substituting the values,

  $\begin{array}{c}{v}^2=u^2-2gh\\ v^2={18}^2-2\times 9.8\times 11\\ =324-215.6\\ =108.4\\ v=10.41\text{m}/\text{s}\end{array}$

b.

Time taken to reach the given height.

t= 2.9 sec and 0.77sec

Given:

Initial speed is 18m/sec vertically upwards.

Height is 11 m.

Formula used:

  $h=ut-\frac{1}{2}g{t}^2$ here h is height, u is initial velocity, g is acceleration due to gravity and t is time.

Calculation:

Given that h =11m, initial velocity is 18.0 m/sec and g is 9.8m/sec² Putting these values in the equation.

  $11=18\times t-0.5\times 9.8\times t^2$

  $4.9t^{{}^2}-18t+11=0$

  $t=\frac{-b\pm \sqrt{(b^2-4ac)}}{2a}$

  $t=\frac{18\pm \sqrt{({18}^2-4\times 4.9\times 11)}}{2\times 4.9}$

  $t=\frac{18\pm \sqrt{({324}^{}-215.6)}}{9.8}$

  $t=\frac{18\pm \sqrt{108.4}}{9.8}$

  $\begin{array}{c}t=\frac{18\pm 10.41}{9.8}\\ =2.9,0.77\sec \end{array}$

c.

To explain: The reason for two answers obtained in part (b).

Given:

Initial speed is 18m/sec vertically upwards.

Height is 11 m.

Explanation:

From part (b), time taken to reach 11 m height is 0.77 sec and 2.9 sec.

There are two answers of part b. One value of t represents the time when it reaches to a height of 11m while going upward and other value t represents the time when it goes to the highest point and then comes downward at a height of 11m from the ground.