Chapter 3, Problem 18P

(a)

To determine

Toprove: The maximum height of a projectile is,

  $h_{\max }=\frac{{\left(v_{0,y}\right)}^2}{2g}$

Answer to Problem 18P

  $h_{\max }=\frac{{\left(v_{0,y}\right)}^2}{2g}$

Explanation of Solution

Introduction:

When a projectile motion happens at an angle ? from the horizontal, the initial velocity of the projectile will have a vertical and a horizontal component.

  $\begin{array}{l}{v}_{0x}=v_0\cdot cos(\theta )\\ v_{0}y=v_0\cdot sin(\theta )\end{array}$

To determine the maximum height reached by a projectile during its flight, the vertical component of its motion is needed.

  

The movement of the projectile is influenced by gravity, vertically. This implies that the vertical portion of the initial velocity will be zero at maximum height.

On its way to maximum height, the projectile must decelerate, come to a complete stop at maximum height, and begin its free fall downward to  the earth. The vertical portion of its initial velocity is,

  $v_{h\max }{}^2=v_{0y}{}^2-2\cdot g\cdot h_{max}$

  0

This equals to,

  $v_{0,y}{}^2=2\cdot g\cdot h_{max}$

Hence, the maximum height reached by the projectile will be,

  $\begin{array}{l}{h}_{max}=\frac{v_{0,y}{}^2}{2.g}=\frac{{(v_0\cdot sin(\theta ))}^2}{2g}\\ h_{max}=\frac{v_0{}^2\cdot si{n}^2(\theta )}{2.g}\end{array}$

Conclusion:

The maximum height of a projectile is,

  $h_{\max }=\frac{{\left(v_{0,y}\right)}^2}{2g}$

(b)

To determine

To prove: The time it takes a projectile to reach its maximum height is,

  $h_{\max }=\frac{v_{0,y}}{g}$

Answer to Problem 18P

  $t=\frac{v_{0,y}}{g}$

Explanation of Solution

Introduction:

The time it takes for a projectile to be projected from a point and landed is named as flight time. This depends on the projectile’s initial velocity and projection angle. The optimum projectile height is if the projectile exceeds a vertical velocity of zero.

In a projectile motion to determine how long an object will remain airborne ,the vertical and horizontal components of the flight path should be considered separately.

The time will take to reach its maximum height in a projectile motion.

  $v_f = v_i + at$

Here, $v_f =0m{s}^{-1}$ and $a =–g =–9.8{\text{ms}}^{\text{-2}}$

Hence,

  $\begin{array}{c}0= v_{0,y} – gt\\ gt= v_{0,y}\\ t=\frac{v_{0,y}}{g}\end{array}$

Conclusion: The time takes a projectile to reach its maximum height is,

  $h_{\max }=\frac{v_{0,y}}{g}$

(c)

To determine

To prove: The time it takes for a projectile to descend from its maximum height to its original elevation is the same as the time to ascend ,which is, $t=\frac{v_{0,y}}{g}$

Answer to Problem 18P

Time of ascent = Time of descent = $\frac{v_{0,y}}{g}$

Explanation of Solution

Introduction:The time that takes a projectile to descend from its maximum height to its original elevation is the same as the time to ascend.

  

The components of a projectile to ascend to its maximum height are as follows.

  $\begin{array}{c}u=u_0\\ v=0\\ a=-g\\ t=?\end{array}$

The time taken to reach its maximum is,

  $\begin{array}{c}v=u+at\\ 0=u_0+\left(-gt\right)\\ gt=u_0\\ t=\frac{u_0}{g}\end{array}$

The components of a projectile to descend from its maximum height are as follows.

  $\begin{array}{c}u=0\\ v=v_f\\ a=g\\ t=?\end{array}$

The time taken to descend from its maximum is,

  $\begin{array}{c}v=u+at\\ v_f=0+gt\\ gt=u_0\\ t=\frac{v_f}{g}\end{array}$

Since $v_f=u_0$ the equation can be rewrited as below.

  $t=\frac{u_0}{g}$

Conclusion:

Time of ascent = Time of descent

(d)

To determine

To prove: The y component of velocity is reversed when a projectile descend to its original elevation $u_y=-v_{y,0}$

Answer to Problem 18P

  $u_y=-v_{y,0}$

Explanation of Solution

Introduction:The x component is the projectile’s horizontal motion and the y component represents the projectile’s vertical motion. The units for expressing the distance horizontal and vertical are meters $\left(m\right)$ . The velocities horizontal and vertical are measured in meters per second $m{s}^{-1}$ .

The time interval between the projectile that reaches the point and is at the maximum height is the same. The magnitude of the velocity on the upward and downward motion at the same point will be the same; the direction will be reversed.

  $\begin{array}{l}\text{For}\text{ascending,}\\ v=0\\ u=?\\ s=h\\ a=-g\\ v^2=u^2+2as\\ 0=u^2-2as\\ u=\sqrt{2gh}\end{array}$

  $\begin{array}{l}\text{For}\text{descending}\\ u=0\\ v=?\\ s=h\\ a=g\\ v^2=u^2+2as\\ v^2=0+2as\\ v^2=\sqrt{2gh}\end{array}$

Hence, the y component of velocity is equals when a projectile descends to its original elevation. But its reverse in sign.

  $u_y=-v_{y,0}$

Conclusion:

The y component of velocity is equals when a projectile descends to its original elevation. But its reverse in sign.

  $u_y=-v_{y,0}$