Chapter 4, Problem 26P

(a)

To determine

The distance covered by the ball when they hit directly upward, and the time to which the ball stays in the air.

Answer to Problem 26P

The distance covered by the ball when they hit directly upward is $\underset{\_}{100\text{m}}$, and the time to which the ball stays in the air is $\underset{\_}{9.2\text{s}}$.

Explanation of Solution

Write the expression for time of projectile to which it reaches the maximum height.

    $t_{\text{top}}=\frac{v_0\sin \theta }{g}$        (I)

Here, $v_0$ is the initial velocity, $g$ is the acceleration due to gravity, and $\theta$ is the angle of projection.

Write the expression for maximum vertical distance covered by the ball from the ground level.

    $y_{\text{top}}=h+v_0\left(\sin \theta \right)t-\frac{1}{2}g{t}^2$        (II)

Here, $h$ is the initial height.

Use equation (I) in (II).

    $\begin{array}{c}{y}_{\text{top}}=h+v_0\left(\sin \theta \right)\frac{v_0\sin \theta }{g}-\frac{1}{2}g{\left(\frac{v_0\sin \theta }{g}\right)}^2\\ =h+\frac{v_0^2{\sin }^2\theta }{g}-\frac{1}{2}\frac{{\left(v_0\sin \theta \right)}^2}{g}\end{array}$        (III)

Write the expression for end of trajectory.

    $\begin{array}{c}{y}_{\text{lands}}=0\\ =h+v_0\left(\sin \theta \right)t_{\text{lands}}-\frac{1}{2}g{t}_{{}_{\text{lands}}}^2\end{array}$        (IV)

Here, $t_{\text{lands}}$ is the time to which the ball stays in the air.

Solve the equation (IV).

    $t_{\text{lands}}=\frac{-\left(v_0\sin \theta \right)\pm \sqrt{{\left(v_0\sin \theta \right)}^2-4\left(-\frac{1}{2}\right)gh}}{-g}$        (V)

Conclusion:

Substitute, $90°$ for $\theta$, $1\text{m}$ for $h$, $45\text{m}/\text{s}$ for $v_0$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in the equation (III), to get $y_{\text{top}}$.

    $\begin{array}{c}{y}_{\text{top}}=1\text{m}+\frac{{\left(45\text{m}/\text{s}\right)}^2{\sin }^{2}90°}{9.8\text{m}/{\text{s}}^2}-\frac{1}{2}\frac{{\left(45\text{m}/\text{s}\times \sin 90°\right)}^2}{9.8\text{m}/{\text{s}}^2}\\ =100\text{m}\end{array}$

Substitute, $90°$ for $\theta$, $1\text{m}$ for $h$, $45\text{m}/\text{s}$ for $v_0$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in the equation (V), to get $t_{\text{lands}}$.

    $\begin{array}{c}{t}_{\text{lands}}=\frac{-\left(45\text{m}/\text{s}\times \sin 90°\right)\pm \sqrt{{\left(45\text{m}/\text{s}\times \sin 90°\right)}^2-4\left(-\frac{1}{2}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(1\text{m}\right)}}{-9.8\text{m}/{\text{s}}^2}\\ =9.2\text{s}\end{array}$

Therefore, the distance covered by the ball when they hit directly upward is $\underset{\_}{100\text{m}}$, and the time to which the ball stays in the air is $\underset{\_}{9.2\text{s}}$.

(b)

To determine

The distance covered by the ball when they hit at an angle of $70°$ with respect to the horizontal, and the time to which the ball stays in the air.

Answer to Problem 26P

The distance covered by the ball when they hit at an angle of $70°$ with respect to the horizontal is $\underset{\_}{92\text{m}}$, and the time to which the ball stays in the air is $\underset{\_}{8.7\text{s}}$.

Explanation of Solution

Conclusion:

Substitute, $70°$ for $\theta$, $1\text{m}$ for $h$, $45\text{m}/\text{s}$ for $v_0$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in the equation (III), to get $y_{\text{top}}$.

    $\begin{array}{c}{y}_{\text{top}}=1\text{m}+\frac{{\left(45\text{m}/\text{s}\right)}^2{\sin }^{2}70°}{9.8\text{m}/{\text{s}}^2}-\frac{1}{2}\frac{{\left(45\text{m}/\text{s}\times \sin 70°\right)}^2}{9.8\text{m}/{\text{s}}^2}\\ =92\text{m}\end{array}$

Substitute, $70°$ for $\theta$, $1\text{m}$ for $h$, $45\text{m}/\text{s}$ for $v_0$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in the equation (V), to get $t_{\text{lands}}$.

    $\begin{array}{c}{t}_{\text{lands}}=\frac{-\left(45\text{m}/\text{s}\times \sin 70°\right)\pm \sqrt{{\left(45\text{m}/\text{s}\times \sin 70°\right)}^2-4\left(-\frac{1}{2}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(1\text{m}\right)}}{-9.8\text{m}/{\text{s}}^2}\\ =8.7\text{s}\end{array}$

Therefore, the distance covered by the ball when they hit at an angle of $70°$ with respect to the horizontal is $\underset{\_}{92\text{m}}$, and the time to which the ball stays in the air is $\underset{\_}{8.7\text{s}}$.