Chapter 11.3, Problem 10P

a.

To determine

To calculate: The height of after 45^th second.

Answer to Problem 10P

Projectile rise 14432 ft. at 45^th second.

Explanation of Solution

Given information: The height of projectile in first second = 15840 ft.

The height of projectile in 2^nd second = 15808 ft.

The height of projectile in third second = 15776

Since, $15840,15808,15776,\mathrm{........}$

Hence, the above series are in A.P

The first term, $a=15840$

Second terms, $a_1=15808$

The common difference, $d=a_1-a=15840-15808=-32$

Formula used: $a_n=a+\left(n-1\right)d$, where n is numbers of terms

Now, 45^th second

Calculation:

  $\begin{array}{l}\therefore a_{45}=15840+\left(45-1\right)\left(-32\right)\\ =15840+44\left(-32\right)\\ =15840-1408\\ =14432\text{ ft}\text{.}\end{array}$

Hence, projectile rise 14432 ft. at 45^th second.

b.

To determine

To calculate: Direction of the projectile and distance of the projectile.

Answer to Problem 10P

The direction of the projectile is downwards and the distance is 3328ft.

Explanation of Solution

Given information: The height of projectile in first second = 15840 ft.

The height of projectile in 2^nd second = 15808 ft.

The height of projectile in third second = 15776

Since, $15840,15808,15776,\mathrm{........}$

Hence, the above series are in A.P

The first term, $a=15840$

Second terms, $a_1=15808$

The common difference, $d=a_1-a=15840-15808=-32$

Formula used: $a_n=a+\left(n-1\right)d$, where n is numbers of terms

Now, 10^th min or 600 second

Calculation:

  $\begin{array}{l}\therefore a_{600}=15840+\left(600-1\right)\left(-32\right)\\ =15840+599\left(-32\right)\\ =15840-19168\\ =-3328\text{ ft}\text{.}\end{array}$

Hence, the negative sign shows the direction of the projectile is downwards and the distance is 3328ft.