Question

Trajectory with a sloped landing Assume an object is launched from the origin with an initial speed \(\left|\mathbf{v}_{0}\right|\) at an angle \(\alpha\) to the horizontal, where \(0 < \alpha < \frac{\pi}{2}\) a. Find the time of flight, range, and maximum height (relative to the launch point) of the trajectory if the ground slopes downward at a constant angle of \(\theta\) from the launch site, where \(0 < \theta < \frac{\pi}{2}\) b. Find the time of flight, range, and maximum height of the trajectory if the ground slopes upward at a constant angle of \(\theta\) from the launch site.

Short answer

Question: Calculate the time of flight, range, and maximum height for the trajectories of a launched object when the ground slopes downward (a) and upward (b) at a constant angle of θ. Answer: 1. For downward-sloping ground: - Time of flight: \(t_d = \frac{2\left|\mathbf{v}_{0}\right|(\sin(\alpha) + \cos(\alpha)\tan(\theta))}{g}\) - Range: \(R_d = \left|\mathbf{v}_{0}\right| \cos(\alpha) t_d\) - Maximum height: \(H_d = \left|\mathbf{v}_{0}\right| \sin(\alpha) t_{max} - \frac{1}{2}gt_{max}^2\) 2. For upward-sloping ground: - Time of flight: \(t_u = \frac{2\left|\mathbf{v}_{0}\right|(\sin(\alpha) - \cos(\alpha)\tan(\theta))}{g}\) - Range: \(R_u = \left|\mathbf{v}_{0}\right| \cos(\alpha) t_u\) - Maximum height: \( H_u = \left|\mathbf{v}_{0}\right| \sin(\alpha) t_{max} - \frac{1}{2}gt_{max}^2 - R_u \tan(\theta)\)

Step-by-step solution

Horizontal and Vertical Displacements

Let's find the expressions for the horizontal and vertical displacements as a function of time, using the given initial speed \(\left|\mathbf{v}_{0}\right|\) and initial angle \(\alpha\). \(x(t) = \left|\mathbf{v}_{0}\right| \cos(\alpha) t\) \(y(t) = \left|\mathbf{v}_{0}\right| \sin(\alpha) t - \frac{1}{2}gt^2\)

Downward-sloping Ground Displacement

Now, let's find the equation for the object's displacement when the ground slopes downward. Downward-sloped ground is described by the equation: \(y_d(x) = -x\tan(\theta)\) In our case, we need to find the time when the object's height coincides with the downward-sloping ground, \(y(t) = y_d(x(t))\): \(\left|\mathbf{v}_{0}\right| \sin(\alpha) t - \frac{1}{2}gt^2 = -\left|\mathbf{v}_{0}\right| \cos(\alpha) t \tan(\theta)\)

Time of Flight for Downward-sloping Ground

Let's solve this equation for the time of flight for the downward-sloping ground case. \(t_d = \frac{2\left|\mathbf{v}_{0}\right|(\sin(\alpha) + \cos(\alpha)\tan(\theta))}{g}\)

Range and Maximum Height for Downward-sloping Ground

Now that we have the time of flight, we can find the range and maximum height for the downward-sloping ground case. Range: \(R_d = \left|\mathbf{v}_{0}\right| \cos(\alpha) t_d\) Maximum height is determined by finding the time when the vertical velocity becomes zero: \(t_{max} = \frac{\left|\mathbf{v}_{0}\right| \sin(\alpha)}{g}\) Then, using this time to find the maximum height: \(H_d = \left|\mathbf{v}_{0}\right| \sin(\alpha) t_{max} - \frac{1}{2}gt_{max}^2\)

Upward-sloping Ground Displacement

Similarly, let's find the equation for the object's displacement when the ground slopes upward. Upward-sloped ground is described by the equation \(y_u(x) = x \tan(\theta)\) In our case, we need to find the time when the object's height coincides with the upward-sloping ground, \(y(t) = y_u(x(t))\): \(\left|\mathbf{v}_{0}\right| \sin(\alpha) t - \frac{1}{2}gt^2 = \left|\mathbf{v}_{0}\right| \cos(\alpha) t \tan(\theta)\)

Time of Flight for Upward-sloping Ground

Let's solve this equation for the time of flight for the upward-sloping ground case. \(t_u = \frac{2\left|\mathbf{v}_{0}\right|(\sin(\alpha) - \cos(\alpha)\tan(\theta))}{g}\)

Range and Maximum Height for Upward-sloping Ground

Now that we have the time of flight, we can find the range and maximum height for the upward-sloping ground case. Range: \(R_u = \left|\mathbf{v}_{0}\right| \cos(\alpha) t_u\) Using \(t_{max}\) from Step 4, we can find the maximum height above the sloped ground: \(H_u = \left|\mathbf{v}_{0}\right| \sin(\alpha) t_{max} - \frac{1}{2}gt_{max}^2 - R_u \tan(\theta)\) These are the time of flight, range, and maximum height for the trajectories of the launched object when the ground slopes downward (a) and upward (b) at a constant angle of \(\theta\).

Key concepts

Time of Flight

Time of flight refers to the total duration that a projectile is airborne. For projectile motion, the time of flight is primarily influenced by the initial speed, the launch angle, and the sloping angle of the ground.

• **Downward-sloping ground**: When the ground slopes downward, the projectile tends to remain in the air longer. The equation to determine the time of flight is: \[t_d = \frac{2\left|\mathbf{v}_0\right| (\sin(\alpha) + \cos(\alpha)\tan(\theta))}{g}\]This reflects how both the upward initial velocity component and the added distance from the downward slope affect the total time in the air.
• **Upward-sloping ground**: On the other hand, when the ground slopes upward, the time of flight generally decreases. The formula used here is: \[t_u = \frac{2\left|\mathbf{v}_0\right| (\sin(\alpha) - \cos(\alpha)\tan(\theta))}{g}\]Here, the slope of the ground works against the projectile, thus reducing airborne time.

Range of Projectile

The range of a projectile is the horizontal distance it covers from the launch point until it lands. This aspect is crucial in determining how far a projectile can travel in a given environment.

• **Downward-sloping ground**: The downward slope works in favor of the projectile, permitting it to cover more distance. Using the time of flight equation for downward slopes, the range can be expressed as:\[R_d = \left|\mathbf{v}_{0}\right| \cos(\alpha) t_d\]The higher value of \(t_d\) due to the slope increases the range significantly compared to flat terrain.
• **Upward-sloping ground**: Conversely, the upward slope restricts the range. This is calculated by:\[R_u = \left|\mathbf{v}_{0}\right| \cos(\alpha) t_u\]The shortened time of flight \(t_u\) results in shorter horizontal travel, making the range less than if the projectile were launched on level ground.

Maximum Height

The maximum height of a projectile is the highest vertical point it reaches in its trajectory. Understanding maximum height allows us to determine how high a projectile will rise and any potential obstacles it might encounter.

• **Calculating maximum height**: The maximum height doesn't directly depend on ground slope. Instead, it is found by analyzing vertical motion alone. The formula is:\[t_{max} = \frac{\left|\mathbf{v}_{0}\right| \sin(\alpha)}{g}\]Solving for height gives:\[H = \left|\mathbf{v}_{0}\right| \sin(\alpha) t_{max} - \frac{1}{2}g t_{max}^2\]
• **Effect of downward/upward slopes**: While the maximum height (relative to the launch point) doesn't change with ground slope, the difference in the ground level can affect relative height expectations. For upward slopes, the relative height is lower due to the ground incline, whereas downward slopes result in a seemingly higher relative height.Effectively, no matter the ground slope, the peak height reached above the launch point remains the same since it only depends on initial upward velocity and gravity.