Question

In: Physics

Find the time th it takes the projectile to reach its maximum height H. Express th in terms of v0, theta, and g (the magnitude of the acceleration due to gravity).

Find the time th it takes the projectile to reach its maximum height H. Express th in terms of v0, theta, and g (the magnitude of the acceleration due to gravity). Find tR, the time at which the projectile hits the ground after having traveled through a horizontal distance R. Express the time in terms of v0, theta, and g. Find H, the maximum height attained by the projectile. Express the time in terms of v0, theta, and g. Find the total distance R (often called the range) traveled in the x direction; see the figure in the problem introduction.

 

Solutions

Expert Solution

Concepts and reason The concept required to solve the given problem is kinematic equations. Initially, find the time taken by the projectile to reach the maximum height by using the equation of motion for a projectile in the vertical direction. Then find the maximum height by applying the equation of motion for a projectile in the vertical direction. Finally, find the horizontal range of the projectile by using the equation of motion for a projectile in the horizontal direction.

Fundamentals

The equation of motion for the projectile in the vertical direction is as follows:

\(v_{\mathrm{fy}}-v_{0 \mathrm{y}}=a t\)

Here, \(v_{\mathrm{fy}}\) is the final velocity in the y-direction, \(v_{0 \mathrm{y}}\) initial velocity in the vertical direction, \(\mathrm{a}\) is the acceleration, and \(\mathrm{t}\) is the time taken to reach the maximum height. The equation of motion for the particle in the vertical direction is as follows:

\(y=v_{0 \mathrm{y}} t_{\mathrm{R}}+\frac{1}{2} a t_{\mathrm{R}}^{2}\)

Here, \(y\) is the vertical distance traveled, and \(t_{\mathrm{R}}\) is when the projectile hits the ground. The equation of motion for the pellet in the vertical direction is, \(v_{\mathrm{fy}}^{2}-v_{0 \mathrm{y}}^{2}=2 a y\)

Here, \(\mathrm{y}\) is the vertical distance covered by the projectile. The equation of motion of the projectile in the horizontal direction is as follows:

\(x=v_{0 \mathrm{x}} t_{\mathrm{R}}+\frac{1}{2} a_{\mathrm{x} 0} t_{\mathrm{R}}^{2}\)

Here, \(x\) is the horizontal distance, and \(a_{0 \mathrm{x}}\) is the acceleration in the horizontal direction.

(A) Substitute 0 for \(v_{\mathrm{fy}}, v_{0} \sin \theta\) for \(v_{0 \mathrm{y}}\), and \(-\mathrm{g}\) for \(\mathrm{a}\) in the equation \(v_{\mathrm{fy}}-v_{0 \mathrm{y}}=a t\) and solve for \(\mathrm{t}\).

$$ \begin{array}{c} 0-v_{0} \sin \theta=-g t \\ t=\frac{v_{0} \sin \theta}{g} \end{array} $$

The acceleration due to gravity on the projectile is taken as negative because when the projectile is thrown upward, it will come down due to gravity acting in the downward direction. The downward direction of the acceleration due to gravity gives the value of equal to \(-g\). At the maximum height of the projectile, the vertical component of velocity vanishes, and only the horizontal component is left.

(B) Substitute 0 for \(\mathrm{y}, v_{0} \sin \theta\) for \(v_{0 \mathrm{y}},-\mathrm{g}\) for an in the equation \(y=v_{0 \mathrm{y}} t_{\mathrm{R}}+\frac{1}{2} a t_{\mathrm{R}}^{2}\), and solve for \(t_{R}\).

$$ \begin{array}{c} 0=\left(v_{0} \sin \theta\right) t_{\mathrm{R}}+\frac{1}{2}(-g) t_{\mathrm{R}}^{2} \\ \left(v_{0} \sin \theta\right) t_{\mathrm{R}}=\frac{1}{2} g t_{\mathrm{R}}^{2} \\ v_{0} \sin \theta=\frac{1}{2} g t_{\mathrm{R}} \\ t_{\mathrm{R}}=\frac{2 v 0 \sin \theta}{g} \end{array} $$

When the projectile reaches the ground, the vertical displacement becomes zero. Therefore, the value of \(y\) is taken as zero.

(C) Substitute 0 for \(v_{\mathrm{fy}}, v_{0} \sin \theta\) for \(v_{0 \mathrm{y}},-\mathrm{g}\) for \(\mathrm{a}\), and \(\mathrm{H}\) for \(\mathrm{y}\) the equation \(v_{\mathrm{fy}}^{2}-v_{0 \mathrm{y}}^{2}=2 a y\) and solve for \(\mathrm{H}\).

$$ \begin{array}{c} 0-\left(v_{0} \sin \theta\right)^{2}=2(-g) H \\ v_{0}^{2} \sin ^{2} \theta=2 g H \\ H=\frac{v_{0}^{2} \sin ^{2} \theta}{2 g} \end{array} $$

At maximum height, the vertical displacement \(y\) is equal to the maximum height of the projectile \(\mathrm{H}\). The vertical component of velocity is zero at maximum height because there is no vertical direction at maximum height.

(D) Substitute \(\mathrm{R}\) for \(\mathrm{x}, v_{0} \cos \theta\) for \(v_{0 \mathrm{x}}, \frac{2 v_{0} \sin \theta}{g}\) for \(t_{\mathrm{R}}\), and \(0 \mathrm{~m} / \mathrm{s}^{2}\) for a in the equation \(x=v_{0 \mathrm{x}} t_{\mathrm{R}}+\frac{1}{2} a_{\mathrm{x} 0} t_{\mathrm{R}}^{2}\), and solve for

R.

$$ \begin{array}{c} R=\left(v_{0} \cos \theta\right)\left(\frac{2 v_{0} \sin \theta}{g}\right)+\frac{1}{2}\left(0 \mathrm{~m} / \mathrm{s}^{2}\right)\left(\frac{2 v_{0} \sin \theta}{g}\right)^{2} \\ =\left(v_{0} \cos \theta\right)\left(\frac{2 v 0 \sin \theta}{g}\right)+0 \\ =\frac{2 v_{0}^{2} \sin \theta \cos \theta}{g} \\ =\frac{v_{0}^{2} \sin 2 \theta}{g} \end{array} $$

The total distance R covered by the projectile is equal to the horizontal displacement of the projectile. The acceleration of the particle in the horizontal direction is equal to zero. Because during the free fall, an object moves only vertically, and there is no component of force parallel to the \(x\) -axis. Therefore, the net acceleration is in the downward direction.


Part A The time taken by the projectile to reach its maximum height is \(\frac{v 0 \sin \theta}{g}\).

Part B The time taken by the projectile when it hits the ground is \(\frac{2 v_{0} \sin \theta}{g}\).

Part C The maximum height attained by the projectile is equal to \(\frac{v_{0}^{2} \sin ^{2} \theta}{2 g}\).

Part D The total distance \(\mathrm{R}\) travelled in the \(\mathrm{x}\) direction is \(\frac{v_{0}^{2} \sin 2 \theta}{g}\).

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