Chapter 1.12, Problem 1AP

A boy $2$ meters tall shoots a toy rocket straight up from head at $10$ meters per second. Assume the acceleration of gravity is $9.8\text{meters}/{\text{sec}}^2$.

(a) What is the highest point above the ground reached by the rocket?

(b) When does the rocket hit the ground?

To determine

(a)

To find:

The highest point above the ground reached by the rocket.

Answer to Problem 1AP

Solution:

The highest point above the ground reached by the rocket is $7.1\text{m}$.

Explanation of Solution

Given:

The height of boy is $2$ meters.

The shootings speed is $10$ meters per seconds.

The acceleration due to gravity is $9.8\text{m}/{\text{s}}^2$.

Formula used:

The equation of motion is,

$y\left(t\right)=v_{0}t+\frac{1}{2}g{t}^2+y_0$

Calculation:

Consider the motion equation.

$y\left(t\right)=v_{0}t+\frac{1}{2}g{t}^2+y_0$… $\left(1\right)$

Differentiate the equation $\left(1\right)$ with respect to $t$.

$\frac{dy}{dt}=v_0+gt$… $\left(2\right)$

At the maximum height the velocity is zero and acceleration due to gravity will be negative because object is moving against the gravity.

Substitute $0$ for $\frac{dy}{dx}$, $10$ meters per sec for $v$ and $-9.8\text{m}/{\text{s}}^2$ for $g$ in equation $\left(2\right)$.

$\begin{array}{rll}10-9.8t& =& 0\\ 9.8t& =& 10\\ t& =& 1.02\text{sec}\end{array}$

Substitute $1.02\text{sec}$ for $t$, $10$ meters per sec for $v$, $2$ for $y_0$ and $-9.8\text{m}/{\text{s}}^2$ for $g$ in equation $\left(1\right)$.

$\begin{array}{rll}y\left(t\right)& =& \left(10\right)1.02+\frac{1}{2}\left(-9.8\right){\left(1.02\right)}^2+2\\ \approx 7.1\text{m}\end{array}$

Therefore, the highest point above the ground reached by the rocket is $7.1\text{m}$.

Conclusion:

Thus, the highest point above the ground reached by the rocket is $7.1\text{m}$.

To determine

(b)

To find:

The time at which the rocket hit the ground.

Answer to Problem 1AP

Solution:

The time at which the rocket hit the ground $1.87\text{sec}$.

Explanation of Solution

Given:

The height of boy is $2$ meters.

The shootings speed is $10$ meters per seconds.

The acceleration due to gravity is $9.8\text{m}/{\text{s}}^2$.

Formula used:

The equation of motion is,

$y\left(t\right)=v_{0}t+\frac{1}{2}g{t}^2+y_0$

Calculation:

Consider the motion equation.

$y\left(t\right)=v_{0}t+\frac{1}{2}g{t}^2+y_0$

At the maximum height the velocity is zero.

$y\left(t\right)=\frac{1}{2}g{t}^2$ …$\left(3\right)$

Substitute $17.29\text{m}$ for $y\left(t\right)$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in equation $\left(3\right)$.

$\begin{array}{rll}17.29& =& \frac{1}{2}\left(9.8\right)t^2\\ t^2& =& 3.52\\ t& =& 1.87\text{sec}\end{array}$

Therefore, the time at which the rocket hit the ground $1.87\text{sec}$.

Conclusion:

Thus, the time at which the rocket hit the ground $1.87\text{sec}$.