Chapter 3, Problem 31P

(a)

To determine

The time taken by the projectile to the point P on the ground.

Answer to Problem 31P

The time taken by the projectile to the point P on the ground is $5.05\text{s}$ .

Explanation of Solution

Given:

The given figure is shown below.

Height of the cliff, $h=125\text{m}$

Formula Used:

The expression to calculate the time taken by the projectile to the point P on the ground is,

  $h=ut+\frac{1}{2}g{t}^2$

Here,

  $u$ is the initial velocity of the projectile and its velocity is zero.

  $g$ is the gravitational acceleration.

  $t$ is the time taken by the projectile.

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}125\text{m}=\left(0\text{m}/\text{s}\right)t+\frac{1}{2}\left(9.8\text{m}/{\text{s}}^2\right)t^2\\ t=\sqrt{\frac{2\left(125\text{m}\right)}{9.8\text{m}/{\text{s}}^2}}\\ t=5.05\text{s}\end{array}$

Conclusion:

Thus, the time taken by the projectile to the point P on the ground is $5.05\text{s}$ .

(b)

To determine

The range of the projectile.

Answer to Problem 31P

The range of the projectile is $262.15\text{m}$ .

Explanation of Solution

Given:

The given figure is shown below.

The initial speed of the projectile is, $u=65\text{m}/\text{s}$ .

The angle of projection of the projectile is $\theta =37°$ .

Formula Used:

Draw the free-body diagram of the system.

The expression to calculate the range of the projectile using speed-distance formula is,

  $X=v_{o}t\cos \theta$

Where,

X is the range of the projectile.

  $v_o$ is the speed of the projectile.

  $t$ is the time taken to reach the projectile.

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}X=\left(65\text{m}/\text{s}\right)\left(5.05\text{}\text{s}\right)\left(\cos 37°\right)\\ =262.15\text{m}\end{array}$

Conclusion:

Thus, the range of the projectile is $262.15\text{m}$ .

(c)

To determine

The horizontal and vertical component of the velocity.

Answer to Problem 31P

The horizontal component of the velocity is $51.91\text{m}/\text{s}$ and the vertical component of the velocity is $39.12\text{m}/\text{s}$ .

Explanation of Solution

Given:

The given figure is shown below.

Formula Used:

Consider the free-body diagram.

The expression to calculate the vertical component of the velocity is,

  $u_y=v_0\sin \theta +gt$

Where,

  $u_y$ is the vertical component of the velocity.

The expression to calculate the horizontal component of the velocity is,

  $u_x=v_0\cos \theta$

Where,

  $u_x$ is the horizontal component of the velocity.

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{u}_y=\left(65\text{}\text{m}/\text{s}\right)\left(\sin 37°\right)+\left(9.8\text{m}/{\text{s}}^2\right)\left(5.05\text{s}\right)\\ =88.61\text{m}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{u}_x=\left(65\text{}\text{m}/\text{s}\right)\left(\cos 37°\right)\\ =51.91\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the horizontal component of the velocity is $51.91\text{m}/\text{s}$ and the vertical component of the velocity is $88.61\text{m}/\text{s}$ .

(d)

To determine

The magnitude of the velocity.

Answer to Problem 31P

The magnitude of the velocity is $102.69\text{m}/\text{s}$ .

Explanation of Solution

Given:

The given figure is shown below.

Formula Used:

From the free body diagram,

The expression to calculate the magnitude of the velocity is,

  $v=\sqrt{u_x^2+u_y^2}$

Where,

  $v$ is the resultant velocity.

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}v=\sqrt{{\left(51.91\text{m}/\text{s}\right)}^2+{\left(88.61\text{m}/\text{s}\right)}^2}\\ =102.69\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the magnitude of the velocity is $102.69\text{m}/\text{s}$ .

(e)

To determine

The angle made by the velocity vector with the horizontal.

Answer to Problem 31P

The angle made by the velocity vector with the horizontal is $59.64°$ .

Explanation of Solution

Given:

The given figure is shown below.

Formula Used:

From the free body diagram,

The expression to calculate the angle made by the velocity vector with the horizontal is,

  $\tan \alpha =\frac{u_y}{u_x}$

Where,

  $\alpha$ is the angle.

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}\tan \alpha =\frac{88.61\text{m}/\text{s}}{51.91\text{m}/\text{s}}\\ \alpha =59.64°\end{array}$

Conclusion:

Thus, the angle made by the velocity vector with the horizontal is $59.64°$ .

(f)

To determine

The maximum height above the cliff top reached by the projectile.

Answer to Problem 31P

The maximum height above the cliff top reached by the projectile is $78.07\text{m}$ .

Explanation of Solution

Given:

The given figure is shown below.

Formula Used:

The expression to calculate the maximum height above the cliff top reached by the projectile is,

  $H=\frac{u^2{\sin }^2\theta }{2g}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}H=\frac{{\left(65\text{m}/\text{s}\right)}^2\left({\sin }^{2}37°\right)}{2\left(9.8\text{m}/{\text{s}}^2\right)}\\ =78.07\text{m}\end{array}$

Conclusion:

Thus, the maximum height above the cliff top reached by the projectile is $78.07\text{m}$ .