Question

kinematics problem set 2 s2 2025 /20 name: rihanna dudu problem 2 (motion 1d and projectile motion no numbers) 2 marks a daredevil is riding on a motorized sled. he is trying to fly using a 45° ramp landing at point 3. he starts from point 1 with a constant acceleration until he reaches point 2 at a height l from the ground. if it took him a time t to cover distance from 1 to 2, derive an expression for horizontal range x in terms of some or all of the given quantities (g, t, l).

Explanation:

Step1: Find initial vertical velocity

The vertical - motion equation for the upward motion from point 2 to the maximum - height and back to point 2 is \(v = v_0+at\). At the maximum - height, the vertical velocity \(v_y = 0\). The time of flight from point 2 to the maximum - height and back to point 2 is \(t\). The acceleration \(a=-g\). The initial vertical velocity \(v_{0y}\) can be related to the time of flight from point 2 to the maximum - height. The time of flight from point 2 to the maximum - height is \(\frac{t}{2}\), so \(0 = v_{0y}-g\frac{t}{2}\), and \(v_{0y}=\frac{gt}{2}\). Also, considering the motion from point 1 to point 2, using \(v_{0y}^2 = 2a\Delta y\). Since the ramp is at \(45^{\circ}\), the initial velocity \(v_0\) has equal horizontal and vertical components. Let the time taken from point 1 to point 2 be \(T\). The vertical displacement from point 1 to point 2 is \(L\). Using \(L = v_{0y}T-\frac{1}{2}gT^{2}\). Since the ramp is at \(45^{\circ}\), the initial velocity \(v_0\) has \(v_{0x}=v_{0y}\).
The time of flight of the projectile motion from point 2 to point 3 is \(t\). The vertical displacement from point 2 to point 3 is \(0\). Using the equation \(y - y_0=v_{0y}t-\frac{1}{2}gt^{2}\), with \(y - y_0 = 0\), we get \(0 = v_{0y}t-\frac{1}{2}gt^{2}\), which gives \(t=\frac{2v_{0y}}{g}\).
The horizontal range \(x = v_{0x}t\). Since the ramp is at \(45^{\circ}\), \(v_{0x}=v_{0y}\).
First, from the motion from point 1 to point 2, using \(L = v_{0y}T-\frac{1}{2}gT^{2}\), we can solve for \(v_{0y}\) as a quadratic equation \(v_{0y}=\frac{gT\pm\sqrt{g^{2}T^{2} + 2gL}}{2}\). We take the positive root \(v_{0y}=\frac{gT+\sqrt{g^{2}T^{2}+2gL}}{2}\) (because the velocity is in the upward - direction).
The time of flight of the projectile motion from point 2 to point 3 is \(t=\frac{2v_{0y}}{g}\).
The horizontal range \(x = v_{0x}t\), and since \(v_{0x}=v_{0y}\), \(x=\frac{(gT + \sqrt{g^{2}T^{2}+2gL})^{2}}{2g}\).

Another approach:

1. First, find the velocity at point 2:

• Using the kinematic equation for vertical motion from point 1 to point 2: \(L = v_{y2}T-\frac{1}{2}gT^{2}\), where \(v_{y2}\) is the vertical velocity at point 2. Solving for \(v_{y2}\) gives \(v_{y2}=\frac{L}{T}+\frac{gT}{2}\).
• Since the ramp is at \(45^{\circ}\), the horizontal velocity at point 2, \(v_{x2}=v_{y2}\).

1. Then, find the time of flight \(t\) for the projectile motion from point 2 to point 3:

• Using the vertical - motion equation \(y - y_0 = v_{y2}t-\frac{1}{2}gt^{2}\), with \(y - y_0 = 0\) (starting and ending at the same height), we have \(0=v_{y2}t-\frac{1}{2}gt^{2}\). Factoring out \(t\) gives \(t(2v_{y2}-gt)=0\). We take \(t=\frac{2v_{y2}}{g}\) ( \(t = 0\) is the starting - time, we ignore it).

1. Finally, find the horizontal range \(x\):

• The horizontal range \(x = v_{x2}t\). Substituting \(v_{x2}=v_{y2}\) and \(t=\frac{2v_{y2}}{g}\), we get \(x=\frac{2v_{y2}^{2}}{g}\).
• Substitute \(v_{y2}=\frac{L}{T}+\frac{gT}{2}\) into the above formula:
• \(x=\frac{2(\frac{L}{T}+\frac{gT}{2})^{2}}{g}=\frac{2(\frac{L^{2}}{T^{2}}+gL+\frac{g^{2}T^{2}}{4})}{g}=\frac{2L^{2}}{gT^{2}} + 2L+\frac{gT^{2}}{2}\)

Answer:

\(x=\frac{2(\frac{L}{T}+\frac{gT}{2})^{2}}{g}=\frac{2L^{2}}{gT^{2}}+2L + \frac{gT^{2}}{2}\)