Chapter 6.1, Problem 27E

a.

To determine

To estimate: velocity of a projectile after 5 sec.

Answer to Problem 27E

Velocity of a projectile after 5 sec is: 240 ft/sec².

Explanation of Solution

Given:

Initial velocity is 400 ft/sec.

  $g=32$ ft/sec²

Formula used:

  $v=u+gt$ ,

where v is the final velocity u is the initial velocity g is the gravitational constant and t is the time.

Calculation:

  $u=400\text{ft/sec}$

  $g=-32{\text{ft/sec}}^{\text{2}}$

  $t=5$

Putting the values in $v=u+gt$

We get,

  $\begin{array}{l}v=400+(-32)\cdot 5\\ v=400-160\\ v=240{\text{ft/sec}}^{\text{2}}\end{array}$

Therefore, the velocity of a projectile after 5 sec is: 240 ft/sec².

b

To determine

To estimate: the lower estimate of height attained after 5 seconds using RRAM_5.

Answer to Problem 27E

Estimated height is: 1520 ft.

Explanation of Solution

Given:

Initial velocity is 400 ft/sec.

  $g=32$ ft/sec²

Formula used:

The area under the velocity curve gives the height (distance in vertical sense) attained by an object.

RRAM₅ refers to Right Rectangle Approximation Method, in which the right points of the each interval are taken as the height of the rectangle to find the area below it and the value of n is 5.

Calculation:

From part (a) the equation for the velocity is given by $v=400-32t$ .

As t is given as 5 sec and n = 5,

  $△x=\frac{b-a}{n}=\frac{5-0}{5}=1$

RRAM₅ method gives

  $\begin{array}{l}RRA{M}_5=△x[y(1)+y(2)+y(3)+y(4)+y(5)]\\ RRA{M}_5=1\left[368+336+304+272+240\right]\\ RRA{M}_5=1520\text{ft}\end{array}$

Therefore, estimated height is: 1520 ft.