Chapter 4, Problem 71P

(a)

To determine

The time taken by the stone to fall to the base of the gorge.

Answer to Problem 71P

The time taken by the stone to fall to the base of the gorge is $\underset{\_}{3.49\text{s}}$.

Explanation of Solution

Write the kinematic equation.

    $y=v_{i}t+\frac{1}{2}a{t}^2$        (I)

Here, $y$ is the distance travelled by the stone in the y direction, $v_i$ is the initial velocity of the stone, $t$ is the time taken by the stone to fall to the base of the gorge, and $a$ is the acceleration of the stone.

Since the initial velocity of the stone is zero, and the acceleration of the stone is equal to the acceleration due to gravity. Thus the above equation is reduced to

    $\begin{array}{c}-y=0+\frac{1}{2}\left(-g\right)t^2\\ t=\sqrt{\frac{2y}{g}}\end{array}$        (II)

Conclusion:

Substitute $60\text{m}$ for $y$, and $9.8\text{m}/{\text{s}}^2$ in equation(II), to find $t$.

    $\begin{array}{c}t=\sqrt{\frac{2\left(60\text{m}\right)}{\left(9.8\text{m}/{\text{s}}^2\right)}}\\ =\text{3}\text{.49}\text{s}\end{array}$

Therefore, the time taken by the stone to fall to the base of the gorge is $\underset{\_}{3.49\text{s}}$.

(b)

To determine

The time taken by the stone to fall to the base of the gorge, if the stone has an initial velocity $20.0\text{m}/{\text{s}}^{\text{2}}$.

Answer to Problem 71P

The time taken by the stone to fall to the base of the gorge, if the stone has an initial velocity $20.0\text{m}/{\text{s}}^{\text{2}}$ is $\underset{\_}{2\text{s}}$.

Explanation of Solution

Write the kinematic equation.

    $y=v_{i}t+\frac{1}{2}a{t}^2$        (III)

Here, $y$ is the distance travelled by the stone in the y direction, $v_i$ is the initial velocity of the stone, $t$ is the time taken by the stone to fall to the base of the gorge, and $a$ is the acceleration of the stone.

Since the initial velocity of the stone is $20.0\text{m}/\text{s}$, the stone travels a distance of $60\text{m}$ in the y direction and the acceleration of the stone is equal to the acceleration due to gravity. Thus the above equation is reduced to

    $-\left(60\text{m}\right)=-\left(20\text{m}/\text{s}\right)t+\frac{1}{2}\left(-9.8\text{m}/{\text{s}}^2\right)t^2$        (IV)

Rearrange the above equation.

    $\left(4.9\text{m}/{\text{s}}^2\right)t^2+\left(20\text{m}/\text{s}\right)t+\left(-60\text{m}\right)=0$

Solve the above quadratic equation.

    $\begin{array}{c}t=\frac{-\left(20\text{m}/\text{s}\right)\pm \sqrt{{\left(20\text{m}/\text{s}\right)}^2-4\left(4.9\text{m}/{\text{s}}^2\right)\left(-60\text{m}\right)}}{2\left(4.9\text{m}/{\text{s}}^2\right)}\\ =\frac{-\left(20\text{m}/\text{s}\right)\pm 39.7\text{m}/\text{s}}{9.8\text{m}/{\text{s}}^2}\end{array}$        (V)

Considering the positive value of time, thus the equation is reduced to

    $\begin{array}{c}t=\frac{-\left(20\text{m}/\text{s}\right)+39.7\text{m}/\text{s}}{9.8\text{m}/{\text{s}}^2}\\ =2\text{s}\end{array}$

Conclusion:

Therefore, the time taken by the stone to fall to the base of the gorge, if the stone has an initial velocity $20.0\text{m}/{\text{s}}^{\text{2}}$ is $\underset{\_}{2\text{s}}$.

(c)

To determine

At what distance the stone will hit the ground from the bridge.

Answer to Problem 71P

The stone will hit the ground at a distance of $\underset{\_}{45.4\text{m}}$ from the bridge.

Explanation of Solution

Given that the stone is projected at an angle of $30°$ from the bridge, and the velocity of projection is $20\text{m}/\text{s}$.

Write the expression for horizontal component of velocity.

    $\begin{array}{c}{v}_{\text{x}}=\left(20\text{m}/\text{s}\right)\text{cos30}°\\ =17.32\text{m}/\text{s}\end{array}$        (VI)

Here, $v_{\text{x}}$ is the horizontal component of velocity.

Write the expression for vertical component of velocity.

    $\begin{array}{c}{v}_{\text{y}}=\left(20\text{m}/\text{s}\right)\sin 30°\\ =10\text{m}/\text{s}\end{array}$        (VII)

Here, $v_{\text{y}}$ is the vertical component of velocity.

Write the kinematic equation.

    $y=v_{y}t+\frac{1}{2}a{t}^2$        (VIII)

Here, $y$ is the distance travelled by the stone in the y direction, $v_y$ is the initial velocity of projection, $t$ is the time taken by the stone to fall to the base of the gorge, and $a$ is the acceleration of the stone.

Since the initial velocity of projection is $10.0\text{m}/\text{s}$, the stone travels a distance of $60\text{m}$ in the y direction and the acceleration of the stone is equal to the acceleration due to gravity. Thus the above equation is reduced to

    $-\left(60\text{m}\right)=-\left(10\text{m}/\text{s}\right)t+\frac{1}{2}\left(-9.8\text{m}/{\text{s}}^2\right)t^2$        (IX)

Rearrange the above equation.

    $\left(4.9\text{m}/{\text{s}}^2\right)t^2+\left(20\text{m}/\text{s}\right)t+\left(-60\text{m}\right)=0$

Solve the above quadratic equation.

    $\begin{array}{c}t=\frac{-\left(10\text{m}/\text{s}\right)\pm \sqrt{{\left(10\text{m}/\text{s}\right)}^2-4\left(4.9\text{m}/{\text{s}}^2\right)\left(-60\text{m}\right)}}{2\left(4.9\text{m}/{\text{s}}^2\right)}\\ =\frac{-\left(10\text{m}/\text{s}\right)\pm 35.7\text{m}/\text{s}}{9.8\text{m}/{\text{s}}^2}\end{array}$        (X)

Considering the positive value of time, thus the equation is reduced to

    $\begin{array}{c}t=\frac{-\left(10\text{m}/\text{s}\right)+35.7\text{m}/\text{s}}{9.8\text{m}/{\text{s}}^2}\\ =2.62\text{s}\end{array}$

Write the expression for horizontal distance travelled by the stone.

    $x=v_{\text{x}}t$        (XI)

Here, $x$ is the horizontal distance travelled by the stone.

Conclusion:

Substitute $17.32\text{m}/\text{s}$ for $v_{\text{x}}$, and $2.62\text{s}$ for $t$ in equation (XI), to find $x$.

    $\begin{array}{c}x=\left(17.32\text{m}/\text{s}\right)\left(2.62\text{s}\right)\\ =45.4\text{m}\end{array}$

Therefore, the stone will hit the ground at a distance of $\underset{\_}{45.4\text{m}}$ from the bridge.