Chapter 3, Problem 102P

(a)

To determine

The value of angle formed by initial velocity vector with horizontal.

Answer to Problem 102P

  ${\theta }_o=86.38°$

Explanation of Solution

Given data:

Final height of ball just above the plate $=0.7\text{}\text{ m}$

Range $=18.4\text{ m}$

Initial speed of ball $=37.5\text{ m/s}$

Formula used:

Vertical position of ball in terms of time $t$ .

  $y=y_o+\left(v_o\sin {\theta }_o\right)t-\frac{1}{2}g{t}^2$

Calculation:

The horizontal distance covered is given as

  $x=\left(v_o\cos {\theta }_o\right)t\mathrm{...}\left(1\right)$

The vertical position can be written as

  $y=y_o+\left(v_o\sin {\theta }_o\right)t-\frac{1}{2}g{t}^2$

From equation (1), put $t=\frac{x}{v_o\cos {\theta }_o}$

  $\begin{array}{l}y=y_o+\left(v_o\sin {\theta }_o\right)t-\frac{1}{2}g{t}^2\\ y=y_o+\left(v_o\sin {\theta }_o\right)\frac{x}{v_o\cos {\theta }_o}-\frac{1}{2}g{\left(\frac{x}{v_o\cos {\theta }_o}\right)}^2\\ y=y_o-\left(\tan {\theta }_o\right)x-\frac{g}{2v_o^2{\cos }^2{\theta }_o}{x}^2\mathrm{...}\left(2\right)\end{array}$

Now, substitute the values of respective parameters in equation

  $\begin{array}{l}y=y_o-\left(\tan {\theta }_o\right)x-\frac{g}{2v_o^2{\cos }^2{\theta }_o}{x}^2\\ 0.7=2.0-\left(\tan {\theta }_o\right)\times 18.4-\frac{9.81}{2\times {\left(37.5\right)}^2\times {\cos }^2{\theta }_o}{\left(18.4\right)}^2\\ 18.4\times \tan {\theta }_o+\frac{1.181}{{\cos }^2{\theta }_o}=1.3\\ 18.4\times \tan {\theta }_o+\frac{1.181}{{\cos }^2{\theta }_o}=1.3\\ 18.4\times \sqrt{{\sec }^2{\theta }_o-1}+\frac{1.181}{{\cos }^2{\theta }_o}=1.3\\ 18.4\times \sqrt{{\frac{1}{{\cos }^2\theta }}_o-1}+\frac{1.181}{{\cos }^2{\theta }_o}=1.3\\ 18.4\times \sqrt{{\frac{1}{{\cos }^2\theta }}_o-1}=1.3-\frac{1.181}{{\cos }^2{\theta }_o}\end{array}$

By squaring both sides,

  $\begin{array}{l}{\left(18.4\times \sqrt{{\frac{1}{{\cos }^2\theta }}_o-1}\right)}^2={\left(1.3-\frac{1.181}{{\cos }^2{\theta }_o}\right)}^2\\ {\frac{338.56}{{\cos }^2\theta }}_o-338.56=1.69+\frac{1.394}{{\cos }^4{\theta }_o}-\frac{3.07}{{\cos }^2{\theta }_o}\\ \frac{1.394}{{\cos }^4{\theta }_o}-\frac{341.63}{{\cos }^2{\theta }_o}+340.25=0\end{array}$

After solving the above equation, the value of ${\theta }_o$

  $\begin{array}{l}\frac{1.394}{{\cos }^4{\theta }_o}-\frac{341.63}{{\cos }^2{\theta }_o}+340.25=0\\ \frac{1}{{\cos }^2{\theta }_o}=\frac{-\left(-341.63\right)\pm \sqrt{{\left(-341.63\right)}^2-4\times 1.394\times 340.25}}{2\times 1.394}\\ \frac{1}{{\cos }^2{\theta }_o}=\frac{341.63\pm \sqrt{116711.0569-1897.234}}{2.7}\\ \frac{1}{{\cos }^2{\theta }_o}=\frac{341.63\pm 338.84}{2.7}\end{array}$

By taking the positive sign

  $\begin{array}{l}\frac{1}{{\cos }^2{\theta }_o}=\frac{341.63+338.84}{2.7}\\ \frac{1}{{\cos }^2{\theta }_o}=252.025\\ {\cos }^2{\theta }_o=\frac{1}{252.025}\\ {\cos }^2{\theta }_o=3.967\times {10}^{-3}\\ \cos {\theta }_o=\sqrt{3.967\times {10}^{-3}}\\ \cos {\theta }_o=0.06299\\ {\theta }_o={\cos }^{-1}\left(0.06299\right)\\ {\theta }_o=86.38°\end{array}$

Conclusion:

The velocity and the horizon made an angle is ${\theta }_o=86.38°$

(b)

To determine

The value of speed with which ball crosses plate.

Answer to Problem 102P

  $v=38.88\text{ m/s}$

Explanation of Solution

Given data:

Final height of ball just above the plate $=0.7\text{}\text{ m}$

Range $=18.4\text{ m}$

Initial speed of ball $=37.5\text{ m/s}$

Formula used:

Speed is given by

  $v=\sqrt{v_x^2+v_y^2}$

Calculation:

The horizontal component of velocity is $v_x$

  $v_x=\frac{dx}{dt}=\frac{d}{dt}\left(v_o\cos {\theta }_o\right)t=v_o\cos {\theta }_o$

The vertical component of velocity is $v_y$

  $v_y=\frac{dy}{dt}=\frac{d}{dt}\left(y_o+v_o\sin {\theta }_o\times t-\frac{1}{2}g{t}^2\right)=v_o\sin {\theta }_o-gt$

Substitute these values into speed formula

  $\begin{array}{l}v=\sqrt{{\left(v_o\cos {\theta }_o\right)}^2+{\left(v_o\sin {\theta }_o-gt\right)}^2}\\ v=\sqrt{{\left(v_o\cos {\theta }_o\right)}^2+{\left(v_o\sin {\theta }_o-g\frac{x}{v_o\cos {\theta }_o}\right)}^2}\\ v=\sqrt{{\left(37.5\times \cos 86.38°\right)}^2+{\left(37.5\times \sin 86.38°-9.81\times \frac{18.4}{37.5\times \cos 86.38°}\right)}^2}\\ v=\sqrt{5.606+1506.2566}\\ v=38.88\text{ m/s}\end{array}$

Conclusion:

The velocity of the ball over the plate is $38.88\text{ m/s}$ .