Chapter 3, Problem 72AP

A dart gun is fired while being held horizontally at a height of 1.00 m above ground level and while it is at rest relative to the ground. The dart from the gun travels a horizontal distance of 5.00 m. A college student holds the same gun in a horizontal position while sliding down a 45.0° incline at a constant speed of 2.00 m/s. How far w ill the dart travel if the student fires the gun when it is 1.00 m above the ground?

To determine

The magnitude of distance travelled by the dart.

Answer to Problem 72AP

Solution:

The magnitude of distance travelled by the dart is $4.14\text{m}$.

Explanation of Solution

Given Info:

The initial speed of the dart is zero, the height of the dart from the ground is $1.00\text{m}$, the horizontal distance travelled by the dart is $5.00\text{m}$, the speed of the gun with respect to the ground is $2.00\text{m}/\text{s}$ and the acceleration due to gravity is $9.80\text{m}/{\text{s}}^{\text{2}}$.

Here the variables measured downward is taken as negative and upward is taken as positive.

Write the formula to calculate the time required to travel by the dart when the gun is at rest.

$t=\sqrt{\frac{2\left(-S\right)}{-g}}$

• t is the time taken by the dart
• S is the downward distance travelled by the dart
• g is the acceleration due to gravity

Substitute $1.00\text{m}$ for S and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate t.

$\begin{array}{c}t=\sqrt{\frac{2\left(-1.00\text{m}\right)}{-\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)}}\\ =\sqrt{\frac{2}{9.80}{\text{s}}^{\text{2}}}\\ =0.452\text{s}\end{array}$

Write the formula to calculate the speed of the dart with respect to the gun.

$vd=\frac{x}{t}$

• $vd$ is the velocity of the dart with respect to the ground
• x is the horizontal distance travelled by the dart

Substitute $5.00\text{m}$ for x and $0.452\text{s}$ for t to calculate $vd$.

$\begin{array}{c}vd=\frac{5.00\text{m}}{0.452\text{s}}\\ =11.1\text{m}/\text{s}\end{array}$

Write the formula to calculate the vertical component speed of the dart.

$v_y=-\left(v_g\right)\sin \theta$

• $v_y$ is the vertical component speed of the dart
• $v_g$ is the speed of the gun with respect to earth
• $\theta$ is the angle made by the dart with respect to vertical

Substitute $2.00\text{m}/\text{s}$ for $v_g$ and $45.0°$ for $\theta$ to calculate $v_y$.

$\begin{array}{c}{v}_y=-\left(2.00\text{m}/\text{s}\right)\sin 45.0°\\ =-\left(2.00\text{m}/\text{s}\right)\left(0.707\right)\\ =-1.41\text{m}/\text{s}\end{array}$

Write the formula to calculate the horizontal component speed of the dart.

$v_x=vd+v_g\cos \theta$

• $v_x$ is the horizontal speed of the dart

Substitute $11.1\text{m}/\text{s}$ for $vd$, $45.0°$ for $\theta$ and $2.00\text{m}/\text{s}$ for $v_g$ to calculate $v_x$.

$\begin{array}{c}{v}_x=11.1\text{m}/\text{s}+\left(2.00\text{m}/\text{s}\right)\cos 45.0°\\ =11.1\text{m}/\text{s}+\left(1.41\text{m}/\text{s}\right)\\ =12.5\text{m}/\text{s}\end{array}$

Write the formula to calculate the vertical speed of the dart after dropping $1.00\text{m}$ to the ground.

$v_y{}^{\prime }=-\sqrt{v_y{}^2-2g\left(-S\right)}$

• $v_y{}^{\prime }$ is the vertical speed of the dart after dropping $1.00\text{m}$

Use $1.41\text{m}/\text{s}$ for $v_y$, $1.00\text{m}$ for S and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g in the above expression to calculate $v_y{}^{\prime }$.

$\begin{array}{c}{v}_y{}^{\prime }=-\sqrt{{\left(1.41\text{m}/\text{s}\right)}^2-2\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\left(-1.00\text{m}\right)}\\ =-\sqrt{1.99{{\text{m}}^2/\text{s}}^2+19.6{{\text{m}}^2/\text{s}}^2}\\ =-4.65\text{m}/\text{s}\end{array}$

The initial vertical component speed of the dart is $-1.41\text{m}/\text{s}$ and final vertical component speed of the dart is $-4.65\text{m}/\text{s}$.

Write the formula to calculate the time of flight for the dart to reach the ground after dropping.

$t^{\prime }=\frac{v_y\text{'}-v_y}{-g}$

• $t^{\prime }$ is the time of flight for the dart

Substitute $-1.41\text{m}/\text{s}$ for $v_y$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for g and $-4.65\text{m}/\text{s}$ for $v_y{}^{\prime }$ to calculate $t^{\prime }$.

$\begin{array}{c}{t}^{\prime }=\frac{-4.65\text{m}/\text{s}-\left(-1.41\text{m}/\text{s}\right)}{-9.80\text{m}/{\text{s}}^{\text{2}}}\\ =\frac{-3.24\text{m}/\text{s}}{-9.80\text{m}/{\text{s}}^{\text{2}}}\\ =0.331\text{s}\end{array}$

Write the formula to calculate the horizontal displacement the dart during flight.

$x^{\prime }=v_x{t}^{\prime }$

• $x^{\prime }$ is the horizontal displacement of the dart during flight

Substitute $0.331\text{s}$ for $t^{\prime }$ and $12.5\text{m}/\text{s}$ for $v_x$ to calculate $x^{\prime }$.

$\begin{array}{c}{x}^{\prime }=\left(12.5\text{m}/\text{s}\right)\left(0.331\text{s}\right)\\ =4.14\text{m}\end{array}$

Conclusion:

Therefore, the magnitude of distance travelled by the dart is $4.14\text{m}$