Chapter 2, Problem 72RE

(a).

To determine

To find the time taken the balls to fall the first $160cm$ . Average velocity for the period.

Answer to Problem 72RE

  $t=0.571\sec$ and velocity is $280\text{cm/sec}$

Explanation of Solution

Given:

  $s=490t^2$ .

Concept Used:

Velocity is the derivative of the positive function with respect to time. At time $t$ the velocity is:

  $v\left(t\right)=\frac{ds}{dt}=\underset{\Delta t\to 0}{\lim }\frac{f\left(t+\Delta t\right)-f\left(t\right)}{\Delta t}$ .

Calculation:

Initial velocity $u=0$ , $s=160cm$ , $g=980{\text{cm/s}}^{\text{2}}$

This gives, $t=\sqrt{\frac{160}{490}}=0.571\sec$

It will take $0.571\sec$ to fall the first $160\text{cm}$

Average velocity= $\frac{u+v}{2}$

Also,

  $\begin{array}{c}{v}^2=2gs\\ v=\sqrt{2gs}\\ =\sqrt{2\times 980\times 160}\\ =560\text{cm/s}\end{array}$

Average velocity will be $\frac{0+560}{2}=280\text{cm/sec}$

Conclusion:

  $t=0.571\sec$ and velocity is $280\text{cm/sec}$

b.

To determine

Velocity of ball and acceleration.

Answer to Problem 72RE

Velocity of ball when reached 160 mark is $56\text{0}\text{cm/s}$ , Acceleration is $980{\text{cm/s}}^{\text{2}}$ .

Explanation of Solution

Given:

  $s=490t^2$ .

Concept Used:

Velocity is the derivative of the positive function with respect to time. At time $t$ the velocity is:

  $v\left(t\right)=\frac{ds}{dt}=\underset{\Delta t\to 0}{\lim }\frac{f\left(t+\Delta t\right)-f\left(t\right)}{\Delta t}$ .

Acceleration= $\frac{v^2}{2s}$ .

Calculation:

The velocity of the ball when it reaches 160 marks= $560\text{cm/sec}$

  $\begin{array}{c}\text{Acceleration}=\frac{v^2}{2s}\\ =\frac{{\left(560\right)}^2}{2\times 160}\\ =980{\text{cm/s}}^{\text{2}}\end{array}$

Conclusion:

Velocity of ball when reached 160 marks= $56\text{0}\text{cm/s}$ , Acceleration is $980{\text{cm/s}}^{\text{2}}$ .