Chapter 2, Problem 39SP

A baseball is thrown straight upward with a speed of 30 m/s. (a) How long will it

rise? (b) How high will it rise? (c) How long after it leaves the hand will it return to

the starting point? (d) When will its speed be 16 m/s?

To determine

The time for which the baseball will rise when it is thrown upward with the velocity of $30\text{ m/s}$.

Answer to Problem 39SP

Solution:

$3.1\text{ s}$

Explanation of Solution

Given data:

The initial velocity of the baseball is $30\text{ m/s}$, and it is thrown upward.

Formula used:

The time taken by object to reach the peak altitude is written as (since at peak final velocity becomes 0),

$t_p=\frac{-v_i}{g}$

Here, $v_i$ is the initial velocity and $g$ is the gravitational acceleration.

Sign convention:Upward and rightward directions are considered as positive, and downward and leftward directions are considered as negative.

Explanation:

Understand that if a motion is under gravity, then the acceleration acting on the body is equal to the acceleration due to gravity. The value of the acceleration due to gravity is $9.8{\text{ m/s}}^2$.

Recall the expression for the time taken by the object to reach the peak altitude.

$t_p=\frac{-v_i}{g}$

Substitute $30\text{ m/s}$ for $v_i$ and $-9.81{\text{ m/s}}^2$ for $g$.

$\begin{array}{c}{t}_p=\frac{-30\text{ m/s}}{-9.81{\text{ m/s}}^2}\\ =3.1\text{ s}\end{array}$

The baseball will rise till to its peak altitude. So, the time for this will be equals to the peak time.

Conclusion:

Hence, the time for which the baseball will rise is $3.1\text{ s}$.

To determine

The height for which the baseball will rise if it is thrown upwards with the velocity of $30\text{ m/s}$.

Answer to Problem 39SP

Solution:

$46\text{ m}$

Explanation of Solution

Given data:

The initial velocity of the baseball is $30\text{ m/s}$, thrown upwards.

Formula used:

The expression for the final velocity of the object in uniformly accelerated motion is written as,

$v_f^2=v_i^2+2as$

Here, $v_i$ is the initial velocity, $a$ is acceleration and $s$ is the displacement.

Explanation:

Understand that if a motion is under gravity, then the acceleration acting on the body is equal to the acceleration due to gravity. The value of the acceleration due to gravity is $9.8{\text{ m/s}}^2$.

Recall the expression for the final velocity of the object in uniformly accelerated motion.

$v_f^2=v_i^2+2as$

Rearrange the expression for $s$.

$s=\frac{v_f^2-v_i^2}{2a}$

Substitute $0\text{ m/s}$ for $v_f$ as at peak velocity becomes zero temporarily, $30\text{ m/s}$ for $v_i$ and $-9.81{\text{ m/s}}^2$ for $g$.

$\begin{array}{c}s=\frac{{\left(0\text{ m/s}\right)}^2-{\left(30\text{ m/s}\right)}^2}{2\left(-9.81{\text{ m/s}}^2\right)}\\ =46\text{ m}\end{array}$

Conclusion:

Hence, the baseball will rise up to $46\text{ m}$.

To determine

The time taken by the baseball when it leaves the hand and return back to the starting point if it thrown upwards with the initial speed of $30\text{ m/s}$.

Answer to Problem 39SP

Solution:

$6.1\text{ s}$

Explanation of Solution

Given data:

The initial velocity of the baseball is $30\text{ m/s}$, thrown upward.

Formula used:

The time taken by an object to reach the peak altitude is written as,

$t_p=\frac{-v_i}{g}$

Here, $v_i$ is the initial velocity and $g$ is the gravitational acceleration.

The total time of flight for any object in projectile motion is written as(since the motion will be symmetrical due to the application of same acceleration),

$t_T=2t_p$

Here, $t_p$ is the time taken by object to reach the peak altitude.

Explanation:

Understand that if a motion is under gravity, then the acceleration acting on the body is equal to the acceleration due to gravity. The value of the acceleration due to gravity is $9.8{\text{ m/s}}^2$.

Recall the expression foe time taken by object to reach the peak altitude.

$t_p=\frac{-v_i}{g}$

Substitute $30\text{ m/s}$ for $v_i$ and $-9.81{\text{ m/s}}^2$ for $g$.

$\begin{array}{c}{t}_p=\frac{-30\text{ m/s}}{-9.81{\text{ m/s}}^2}\\ =3.06\text{ s}\end{array}$

Recall the expression for total time of flight for any object in projectile motion.

$t_T=2t_p$

Substitute $3.06\text{ s}$ for $t_p$.

$\begin{array}{c}{t}_T=2\left(3.06\text{ s}\right)\\ =6.1\text{ s}\end{array}$

Conclusion:

Hence, the time taken by the baseball when it leaves the hand and return back to the starting point is $6.1\text{ s}$.

To determine

The time at which the velocity of the baseball will become $16\text{ m/s}$, when it is thrown upwards with the velocity of $30\text{ m/s}$.

Answer to Problem 39SP

Solution:

$1.4\text{ s}$ and $4.7\text{ s}$.

Explanation of Solution

Given Data:

The initial velocity of the baseball is $30\text{ m/s}$, thrown upward.

Formula used:

The expression for the final velocity is written as,

$v_f=v_i+at$

Here, $u$ is the initial velocity and $a$ is the acceleration and $t$ is the time.

Explanation:

Understand that if a motion is under gravity, then the acceleration acting on the body is equal to the acceleration due to gravity. The value of the acceleration due to gravity is $9.8{\text{ m/s}}^2$.

Write the expression for the final velocity of the baseball when first time its velocity become $16\text{ m/s}$.

$v_f=v_i+a{t}_1$

Here, $t_1$ is the time at which first time velocity of baseball will become $16\text{ m/s}$.

Substitute $16\text{ m/s}$ for $v_f$, $-9.81{\text{ m/s}}^2$ for $a$ and $30\text{ m/s}$ for $v_i$.

$\begin{array}{c}16\text{ m/s}=30\text{ m/s}+\left(-9.81{\text{ m/s}}^2\right)t_1\\ t_1=\frac{16\text{ m/s}-30\text{ m/s}}{-9.81{\text{ m/s}}^2}\\ =1.4\text{ s}\end{array}$

This time is found when the baseball moves in upward direction opposite to gravity. So, the value of gravitational acceleration is taken negative.

Write the expression for the final velocity of the baseball when second time its velocity becomes $16\text{ m/s}$(during the downward motion of the ball),

$v_f=v_i+a{t}_2$

Here, $t_2$ is the time at which second time velocity of baseball will become $16\text{ m/s}$.

Substitute $-16\text{ m/s}$ for $v_f$, $9.81{\text{ m/s}}^2$ for $a$ and $0\text{ m/s}$ for $v_i$.

$\begin{array}{c}-16\text{ m/s}=0\text{ m/s}+\left(9.81{\text{ m/s}}^2\right)t_2\\ t_2=\frac{16\text{ m/s}}{9.81{\text{ m/s}}^2}\\ =1.6\text{ s}\end{array}$

This time is found when the baseball moves in downward direction from the top of the height in the direction of gravitational acceleration. So, the value of gravitational acceleration and the final velocity both will be negative.

Write the expression for the total time, from the starting, when the velocity of baseball become $16\text{ m/s}$ for the second time.

$T_2=t_p+t_2$

Substitute $3.1\text{ s}$ for $t_p$ and $1.6\text{ s}$ for $t_2$.

$\begin{array}{c}{T}_2=3.1\text{ s}+1.6\text{ s}\\ =4.7\text{ s}\end{array}$

Add the time taken to reach the maximum height by the baseball and the time when second time its velocity become $16\text{ m/s}$ from the maximum height to calculate the total time when second time velocity of baseball become $16\text{ m/s}$.

Conclusion:

Hence, the values of time at which the velocity of the baseball will become $16\text{ m/s}$ is $1.4\text{ s}$ and $4.7\text{ s}$.