Chapter 3, Problem 30P

(a)

To determine

The maximum height reached by projectile.

Answer to Problem 30P

68.01 m

Explanation of Solution

Given:

The initial speed is 65.2 m/s and the angle is 34.5°.

Formula Used:

The height is,

  $H=\frac{u^2{\sin }^2\theta }{2g}$

Calculation :

  $\begin{array}{l}\text{Initial speed,}u=65.2\text{ m/s}\\ \text{Angle,}\theta \text{ =}\text{34}\text{.5°}\end{array}$

Substituting the given values

  $\begin{array}{l}H=\frac{{\left(65.2\right)}^2{\sin }^2\left(34.5\right)}{2\times 10}\\ H=68.01\text{m}\end{array}$

Conclusion:

The maximum height of projectile is 68.01 m.

(b)

To determine

Total time in air.

Answer to Problem 30P

7.38 sec.

Explanation of Solution

Given:

The initial speed is 65.2 m/s and angle is 34.5°.

Formula Used :

Time of flight is,

  $T=\frac{2u\sin \theta }{g}$

Calculation :

  $\begin{array}{l}\text{Initial speed,}u=65.2\text{ m/s}\\ \text{Angle,}\theta \text{ =}\text{34}\text{.5°}\end{array}$

Substituting the given values,

  $\begin{array}{l}T=\frac{2\times 65.2\sin 34.5}{10}\\ t=7.38\text{ sec}\end{array}$

Conclusion :

Thus, the total time in air is 7.38 sec.

(c)

To determine

To find :Total horizontal distance covered.

Answer to Problem 30P

396.86 m

Explanation of Solution

Given:

  $\begin{array}{l}\text{Initial speed,}u=65.2\text{ m/s}\\ \text{Angle,}\theta \text{ =}\text{34}\text{.5°}\end{array}$

Formula Used :

  $R=\frac{u^2\sin 2\theta }{g}$

Calculation :

Substituting the given values

  $\begin{array}{l}R=\frac{{\left(65.2\right)}^2\sin \left(2\times 34.5\right)}{10}\\ R=396.86\text{m}\end{array}$

Conclusion :

Thus, the horizontal distance covered is 396.86m.

(d)

To determine

The velocity of projectile after the given time.

Answer to Problem 30P

21.92 m/s

Explanation of Solution

Given:

Time, t = 1.50 sec

Formula Used :

  $\begin{array}{l}{v}_y=u_y-at\\ u_y=u\sin \theta \end{array}$

Calculation :

Substituting the values, to find the initial velocity.

  $\begin{array}{l}{u}_y=65.2\sin \left(34.5\right)\\ u_y=36.92\end{array}$

Substituting the value

  $\begin{array}{l}{v}_y=36.92-10\times 1.50\\ v_y=21.92\text{ m/s}\end{array}$

Conclusion :

Thus,the velocity will be 21.92 m/s after time 1.50 sec.