Chapter 13.3, Problem 28E

a.

To determine

To find : The instantaneous velocity of the arrow after one second.

Answer to Problem 28E

The instantaneous velocity of the arrow after one second is 56.34m/s.

Explanation of Solution

Given information:

The arrow is shot upward on the moon with a velocity of 58m/s, its height after t seconds is given by $H=58t-0.83t^2$.

Calculation :

The instantaneous velocity after one second is the derivative of the function $H=58t-0.83t^2$ at $t=1$ that is $H\text{'}\left(1\right)$.

The derivative of a function $f$ at a number $a$ , denoted by $f\text{'}\left(a\right)$ ,is

  $f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$

provided this limit exist.

According to definition, the derivative of $H=58t-0.83t^2$ , with $t=1$ is

  $\begin{array}{l}H\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ H\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{58\left(1+h\right)-0.83{\left(1+h\right)}^2-\left(58\cdot 1-0.83\cdot 1^2\right)}{h}\\ H\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{58+58h-0.83\left(1+2h+h^2\right)-57.17}{h}\\ H\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{58+58h-0.83-1.66h+0.83h^2-57.17}{h}\\ H\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{56.34h+0.83h^2}{h}\\ H\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{\cancel{h}\left(56.34+0.83h\right)}{\cancel{h}}\\ H\text{'}\left(1\right)=\underset{h\to 0}{\lim }\left(56.34+0.83h\right)\\ H\text{'}\left(1\right)=56.34\end{array}$

Hence,

The instantaneous velocity of the arrow after one second is 56.34m/s.

b.

To determine

To find : The instantaneous velocity of the arrow when $t=a$.

Answer to Problem 28E

The instantaneous velocity on the arrow when $t=a$ is $58-1.66a$ m/s.

Explanation of Solution

Given information:

The arrow is shot upward on the moon with a velocity of 58m/s, its height after t seconds is given by $H=58t-0.83t^2$.

Calculation :

The instantaneous velocity at $t=a$ is the derivative of the function $H=58t-0.83t^2$ at $t=a$ that is $H\text{'}\left(a\right)$.

The derivative of a function $f$ at a number $a$ , denoted by $f\text{'}\left(a\right)$ ,is

  $f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$

provided this limit exist.

According to definition, the derivative of $H=58t-0.83t^2$ , with $t=a$ is

  $\begin{array}{l}H\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ H\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{58\left(a+h\right)-0.83{\left(a+h\right)}^2-\left(58a-0.83a^2\right)}{h}\\ H\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{58a+58h-0.83\left(a^2+2ah+h^2\right)-\left(58a-0.83a^2\right)}{h}\\ H\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{58a+58h-0.83a^2-1.66ah-0.83h^2-58a+0.83a^2}{h}\\ H\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{58h-1.66ah-0.83h^2}{h}\\ H\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\cancel{h}\left(58-1.66a-0.83h\right)}{\cancel{h}}\\ H\text{'}\left(a\right)=\underset{h\to 0}{\lim }\left(58-1.66a-0.83h\right)\\ H\text{'}\left(a\right)=58-1.66a\end{array}$

Hence,

The instantaneous velocity on the arrow when $t=a$ is $58-1.66a$ m/s.

c.

To determine

To find : The time t when the arrow hit the moon.

Answer to Problem 28E

Arrow will hit the moon at time $t=\frac{58}{0.83}$ seconds

Explanation of Solution

Given information:

The arrow is shot upward on the moon with a velocity of 58m/s, its height after t seconds is given by $H=58t-0.83t^2$.

Calculation :

The arrow will hit the moon when $H\left(t\right)=0$.

Therefore,

  $\begin{array}{l}H\left(t\right)=0\\ 58t-0.83t^2=0\\ t\left(58-0.83t\right)=0\\ t=0\text{ or }\left(58-0.83t\right)=0\\ t=0\text{ or }-0.83t=-58\\ t=0\text{ or }t=\frac{58}{0.83}\end{array}$

Since, at $t=0$ the arrow is being shot upwards, therefore arrow will hit the moon at time $t=\frac{58}{0.83}$ seconds

Hence,

Arrow will hit the moon at time $t=\frac{58}{0.83}$ seconds

d.

To determine

To find : The velocity with what the arrow will hit the moon.

Answer to Problem 28E

Arrow will hit the moon at time $t=\frac{58}{0.83}$ seconds

Explanation of Solution

Given information:

The arrow is shot upward on the moon with a velocity of 58m/s, its height after t seconds is given by $H=58t-0.83t^2$.

Calculation :

From (b) the instantaneous velocity on the arrow when $t=a$ is $H\text{'}\left(a\right)=58-1.66a$ m/s.

Also, the time at which arrow will hit the moon is $t=\frac{58}{0.83}$ seconds.

To find the velocity with what the arrow will hit the moon, substitute $t=\frac{58}{0.83}$ in above instantaneous velocity and solve:

  $\begin{array}{l}H\text{'}\left(\frac{58}{0.83}\right)=58-1.66\cdot \frac{58}{0.83}\\ H\text{'}\left(\frac{58}{0.83}\right)=58-2\cdot 58\\ H\text{'}\left(\frac{58}{0.83}\right)=-58\end{array}$

Hence,

The velocity at which the arrow will hit the moon is 58 m/s.