Chapter 2, Problem 62P

(a)

To determine

Maximum height that can be reached by stone.

Answer to Problem 62P

Maximum height is $21.1\text{m}$.

Explanation of Solution

Initial speed is $19.6\text{m}/\text{s}$ and the initial height of stone from the ground is $1.50\text{m}$.

Write the Newton’s equation for an object moving vertically upwards.

$v_f^2-v_i^2=-2g\left(y_f-y_i\right)$

Here, the velocity at the highest point of motion is $v_f$, initial velocity of motion is $v_i$, gravitational acceleration is $g$, final position from the ground is $y_f$, and the initial position from the ground is $y_i$.

Conclusion:

Substitute $0\text{m}/\text{s}$ for $v_f$, $19.6\text{m}/\text{s}$ for $v_i$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $1.50\text{m}$ for $y_i$ in the above equation to find $y_f$.

$\begin{array}{c}{\left(0\text{m}/\text{s}\right)}^2-{\left(19.6\text{m}/\text{s}\right)}^2=-2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(y_f-1.50\text{m}\right)\\ 384{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}=19.6\text{m}/{\text{s}}^{\text{2}}\left(y_f-1.50\text{m}\right)\end{array}$

Rewrite the above relation in terms of $y_f$.

$\begin{array}{c}{y}_f=\frac{384{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{19.6\text{m}/{\text{s}}^{\text{2}}}+1.50\text{m}\\ =21.1\text{m}\end{array}$

(b)

To determine

Total time taken by the stone for its whole journey before reaching the ground.

Answer to Problem 62P

Time taken is $4.08\text{s}$.

Explanation of Solution

Initial speed is $19.6\text{m}/\text{s}$ and the initial height of stone from the ground is $1.50\text{m}$.

Write the equation for time taken by the stone to reach the maximum height.

$t_{up}=\sqrt{\frac{2\left(y_f-y_i\right)}{g}}$ (I)

Here, the time taken by stone for its upward motion is $t_{up}$.

Write the equation for time taken by the stone for its downward motion.

$t_{down}=\sqrt{\frac{2y_f}{g}}$ (II)

Here, the time taken by stone for its downward motion is $t_{down}$.

Write the equation for total time taken by stone for its whole travelling before reaching he ground.

$t=t_{up}+t_{down}$ (III)

Here, the total time taken is $t$.

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $21.1\text{m}$ for $y_f$ and $1.50\text{m}$ for $y_i$ in equation (I) to find $t_{up}$.

$\begin{array}{c}{t}_{up}=\sqrt{\frac{2\left(21.1\text{m}-1.50\text{m}\right)}{\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}}\\ =2.0\text{s}\end{array}$

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $21.1\text{m}$ for $y_f$ a in equation (II) to find $t_{down}$.

$\begin{array}{c}{t}_{down}=\sqrt{\frac{2\left(21.1\text{m}\right)}{9.8\text{m}/{\text{s}}^{\text{2}}}}\\ =2.08\text{s}\end{array}$

Substitute $2.0\text{s}$ for $t_{up}$ and $2.08\text{s}$ for $t_{down}$ in equation (III) to find $t$.

$\begin{array}{c}t=2.0\text{s}+2.08\text{s}\\ =\text{4}\text{.08}\text{s}\end{array}$

Therefore, the time taken is $4.08\text{s}$.