Chapter 13.3, Problem 37E

(a)

To determine

To find: The instantaneous velocity of the arrow after 1 second.

Answer to Problem 37E

56.34 m/s

Explanation of Solution

Given: Velocity on the moon If an arrow is shot upwards on the moon with a velocity of 58 m/s, its height ( in meters) after t seconds is given by

  $H=58t-0.83t^2$

The relation of height and time of an arrow.

  $H=58t-0.83t^2$

Differentiate w.r.t t

  $v_{\text{instanraneous}}=\frac{dH}{dt}=58-1.66t$

Put t=1

  $v_{\text{instanraneous}}=58-1.66\left(1\right)=56.34$

Hence, the instantaneous velocity of an arrow after 1 second would be 56.34 m/s.

(b)

To determine

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

Answer to Problem 37E

(58-1. 66a ) m/s

Explanation of Solution

Given: Velocity on the moon If an arrow is shot upwards on the moon with a velocity of 58 m/s, its height ( in meters) after t seconds is given by

  $H=58t-0.83t^2$

The relation of height and time of an arrow.

  $H=58t-0.83t^2$

Differentiate w.r.t t

  $v_{\text{instanraneous}}=\frac{dH}{dt}=58-1.66t$

Put t=a $v_{\text{instanraneous}}=58-1.66a$

Hence, the instantaneous velocity of an arrow after a second would be (58-1. 66a ) m/s.

(c)

To determine

To find: The time when arrow hit the moon.

Answer to Problem 37E

69.88 seconds.

Explanation of Solution

Given: Velocity on the moon If an arrow is shot upwards on the moon with a velocity of 58 m/s, its height ( in meters) after t seconds is given by

  $H=58t-0.83t^2$

The relation of height and time of an arrow.

  $H=58t-0.83t^2$

Put H =0

  $\begin{array}{l}0=58t-0.83t^2\\ t\left(58-0.83t\right)=0\\ t=0,69.88\end{array}$

Hence, After 69.88 seconds arrow will hit the moon.

(d)

To determine

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

Answer to Problem 37E

58 m/s

Explanation of Solution

Given: Velocity on the moon If an arrow is shot upwards on the moon with a velocity of 58 m/s, its height ( in meters) after t seconds is given by

  $H=58t-0.83t^2$

The relation of height and time of an arrow.

  $H=58t-0.83t^2$

Differentiate w.r.t t

  $v_{\text{instanraneous}}=\frac{dH}{dt}=58-1.66t$

Put t=69.88

  $\begin{array}{l}v=58-1.66\left(69.88\right)\\ v=58\end{array}$

Hence, the arrow hit the moon with 58 m/s.