Chapter 2, Problem 45SP

A body is projected downward at an angle of $30.0°$ with the horizontal from the top of a building 170 m high. Its initial speed is 40.0 m/s. (a) How long will it take before

striking the ground? (b) How far from the foot of the building will it strike? (c) At what angle with the horizontal will it strike?

To determine

The time taken by the body to hit the ground if it is projected downward at an angle of $30.0°$ with the horizontal from a height of $170\text{ m}$ with $40.0\text{ m/s}$ initial velocity.

Answer to Problem 45SP

Solution:

$\text{8}\text{.27 s}$

Explanation of Solution

Given data:

The downward angle made by the body with the horizontal is $30.0°$.

The initial speed of body is $40.0\text{ m/s}$.

The height of the building from the ground is $170\text{ m}$.

Formula used:

The expression for the vertical component of the initial velocity is written as,

$v_{iy}=v_i\sin \theta$

Here, $v_i$ is the initial velocity and $\theta$ is the projecting angle.

The expression for the vertical displacement of the object in horizontal motion is written as,

$y=v_{iy}t+\frac{1}{2}{a}_y{t}^2$.

Here, $v_{iy}$ is the $y$ component of the initial velocity, $t$ is the time of flight, and $a_y$ is the acceleration in $y$ direction.

Explanation:

Draw the diagram according to the problem.

Here, $\theta$ is the projecting angle of the body, $y$ is the height from which the body is thrown, $v_i$ is the initial velocity, $v_{ix}$ and $v_{iy}$ are the horizontal and vertical component of the initial velocity respectively, $v_f$ is the final velocity and $v_{fy}$ is the vertical component of it, $x$ is the distance from starting point to striking point, and ${\theta }_1$ is the angle by which body strike on the ground.

Recall the expression for the vertical component of the initial velocity.

$v_{iy}=v_i\sin \theta$

Substitute $30.0°$ for $\theta$ and $40.0\text{ m/s}$ for $v_i$

$\begin{array}{c}{v}_{iy}=\left(40.0\text{ m/s}\right)\sin 30.0°\\ =20\text{ m/s}\end{array}$

Recall the expression for the vertical displacement of the object.

$y=v_{iy}t+\frac{1}{2}{a}_y{t}^2$

Substitute $20\text{ m/s}$ for $v_{iy}$, $-170\text{ m}$ for $y$, $40.0\text{ m/s}$ for $v_i$, and $-9.81{\text{ m/s}}^2$ for $a_y$

$-170\text{ m}=\left(20\text{ m/s}\right)t+\frac{1}{2}\left(-9.81{\text{ m/s}}^2\right)t^2$

Solve the quadratic equation for $t$.

$t=8.27\text{ s}$

Or,

$t=-4.19\text{ s}$

Understand that, the time cannot be negative. So, the time of flight will be $8.27\text{ s}$.

Conclusion:

Hence, the time taken by the body to hit the ground is $8.27\text{ s}$.

To determine

The horizontal distance traveled by the body from the foot of the building to striking point, if it is projected downward from the building of height $170\text{ m}$, at angle of $30.0°$ with the horizontalwith an initial speed of $40.0\text{ m/s}$.

Answer to Problem 45SP

Solution:

$286\text{ m}$

Explanation of Solution

Given data:

The downward angle made by the body with the horizontal is $30.0°$.

The initial speed of body is $40.0\text{ m/s}$.

The height of the building from the ground is $170\text{ m}$.

Formula used:

The expression for the horizontal component of the initial velocity is written as,

$v_{ix}=v_i\cos \theta$

Here, $v_i$ is the initial velocity and $\theta$ is the projecting angle.

The expression for the horizontal displacement of the body which follows projectile motion is written as,

$x=v_{ix}t$ (when no acceleration acts in horizontal direction)

Here, $\theta$ is the projecting angle, $v_{ix}$ is the $x$ component of the initial velocity, and $t$ is the time of flight.

Explanation:

Recall the expression for the horizontal component of the initial velocity.

$v_{ix}=v_i\cos \theta$

Substitute $30.0°$ for $\theta$ and $40.0\text{ m/s}$ for $v_i$

$\begin{array}{c}{v}_{ix}=\left(40.0\text{ m/s}\right)\cos 30.0°\\ =34.64\text{ m/s}\end{array}$

Recall the expression for the horizontal displacement of the body which follows projectile motion.

$x=v_{ix}t$

Substitute $34.64\text{ m/s}$ for $v_{ix}$ and $8.27\text{ s}$ for $t$

$\begin{array}{c}x=\left(34.64\text{ m/s}\right)\left(8.27\text{ s}\right)\\ =286\text{ m}\end{array}$

Conclusion:

Hence, the distance traveled by the body from the foot of the building to striking point is $286\text{ m}$

To determine

The angle made by the body with horizontal when it will strike the groundif it is projected downward at an angle of $30.0°$ with the horizontal from height of $170\text{ m}$ and with $40.0\text{ m/s}$ initial velocity.

Answer to Problem 45SP

Solution:

$60°$

Explanation of Solution

Given data:

The downward angle made by the body with the horizontal is $30.0°$.

The initial speed of body is $40.0\text{ m/s}$.

The height of the building from the ground is $170\text{ m}$.

Formula used:

The expression for the horizontal component of the initial velocity is written as,

$v_{ix}=v_i\cos \theta$

Here, $v_i$ is the initial velocity and $\theta$ is the projecting angle.

The expression for the vertical component of the initial velocity is written as,

$v_{iy}=v_i\sin \theta$

The expression for the $y$ component of the final velocity is written as,

$v_{fy}=v_{iy}+at$

Here, $v_{iy}$ is the $y$ component of the initial velocity, $a$ is the acceleration, and $t$ is the time.

The expression for the $\tan \theta$ is written as,

$\tan \theta =\left|\frac{y}{x}\right|$

Here, $\theta$ is the angle, $y$ is the vertical component of the vector and $x$ is the horizontal component of the vector.

Explanation:

Redraw the diagram according to problem.

Recall the expression for the horizontal component of the initial velocity.

$v_{ix}=v_i\cos \theta$

Substitute $30.0°$ for $\theta$ and $40.0\text{ m/s}$ for $v_i$

$\begin{array}{c}{v}_{ix}=\left(40.0\text{ m/s}\right)\cos 30.0°\\ =34.64\text{ m/s}\end{array}$

Recall the expression for the vertical component of the initial velocity.

$v_{iy}=v_i\sin \theta$

Substitute $30.0°$ for $\theta$ and $40.0\text{ m/s}$ for $v_i$

$\begin{array}{c}{v}_{iy}=\left(40.0\text{ m/s}\right)\sin 30.0°\\ =20\text{ m/s}\end{array}$

Recall the expression for the $y$ component of the final velocity.

$v_{fy}=v_{iy}+at$

Substitute $20\text{ m/s}$ for $v_{iy}$, $-9.81{\text{ m/s}}^2$ for $a$, and $8.27\text{ s}$ for $t$

$\begin{array}{c}{v}_{fy}=20\text{ m/s}+\left(-9.81{\text{ m/s}}^2\right)\left(8.27\text{ s}\right)\\ =-61.13\text{ m/s}\end{array}$

Write the expression for the angle made by the body with the horizontal when it strikes the ground.

$\tan {\theta }_1=\left|\frac{v_{fy}}{v_{ix}}\right|$ (Since no acceleration acts in horizontal direction thus horizontal component of its final velocity will be same as the horizontal component of its initial velocity)

Or,

${\theta }_1={\tan }^{-1}\left|\frac{v_{fy}}{v_{ix}}\right|$

Substitute $-61.13\text{ m/s}$ for $v_{fy}$ and $34.64\text{ m/s}$ for $v_{ix}$.

$\begin{array}{c}{\theta }_1=ta{n}^{-1}\left|\frac{61.13\text{ m/s }}{34.64\text{ m/s}}\right|\\ =60°\end{array}$

Understand that, the velocity along the horizontal remains constant throughout the given projectile motion but the vertical component of the velocity changes.

Conclusion:

Hence, the angle made by the body with horizontal when it will strike is $60°$.