Chapter 3, Problem 74GP

(a)

To determine

The x and y component of the initial velocity.

Answer to Problem 74GP

The x component of the initial velocity is $13.38\text{m}/\text{s}$ and calculate the y component of the initial velocity is $12.04\text{m}/\text{s}$ .

Explanation of Solution

Given info:

The initial velocity of the ball is, $u=18\text{m}/\text{s}$ .

The angle of projection of the ball is, $\theta =42°$ .

Formula Used:

The expression to calculate the x component of the initial velocity is,

  $u_x=u\cos \theta$

The expression to calculate the y component of the initial velocity is,

  $u_y=u\sin \theta$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{u}_x=\left(18\text{m}/\text{s}\right)\left(\cos 42°\right)\\ =13.38\text{m}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{u}_y=\left(18\text{m}/\text{s}\right)\left(\sin 42°\right)\\ =12.04\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the x component of the initial velocity is $13.38\text{m}/\text{s}$ and calculate the y component of the initial velocity is $12.04\text{m}/\text{s}$ .

(b)

To determine

The height at which the ball strikes the building.

Answer to Problem 74GP

The height at which the ball strikes at another building is $33.28\text{m}$ .

Explanation of Solution

Given info:

The distance of the nearby building is, $x=55\text{m}$ .

Formula Used:

The expression to calculate the time in which the ball reaches to nearby building is,

  $t=\frac{x}{u_x}$

The expression to calculate the height at which the ball strikes at another building is,

  $h=-u_{y}t+\frac{1}{2}g{t}^2$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}t=\frac{55\text{m}}{13.38\text{m}/\text{s}}\\ =4.11\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}h=\left(-12.04\text{m}/\text{s}\right)\left(4.11\text{s}\right)+\frac{1}{2}\left(9.8\text{m}/{\text{s}}^2\right){\left(4.11\text{s}\right)}^2\\ =33.28\text{m}\end{array}$

Conclusion:

Thus, the height at which the ball strikes at another building is $33.28\text{m}$ .