Unit 2 - Challenge Question - Math Symbolics

Given:

• Use acceleration of gravity up is negative the acceleration of gravity down which is equal to 10 meters/sec/sec
• The horizontal distance from the kicking point at the top of a hill to the spot where the ball finally impacts further down the hill is $$\Delta{x}$$
• Drawing a horizontal line at the point of kicking:
• The angle of the kick up is $$\alpha_{up}$$
• The hill down is a straight inclined plane
• The angle of the hill down from the horizontal is $$\beta$$
• The initial velocity of the kicked ball is $$V_0$$

Rationale:

• The two given angles are used differently:
• Angle alpha is used to determine the x and y initial velocities
• Transversing angle beta between two parallel lines allows angle beta to be the right most angle of a right triangle. This triangle has a y-axis equal to the hill height andd an x-axis equal to Delta-x.
• Two simultaneous equations of the height of the hill allow handling two unknowns in each of these equations.

Solve: Initial velocities of the kicked ball

• $$\frac{V_{0-y}}{V_0} = \sin{\alpha}$$
• $$V_{0-y} = \sin{\alpha} * V_0$$
• $$\frac{V_{0-x}}{V_0} = \cos{\alpha}$$
• $$V_{0-x} = \cos{\alpha} * V_0$$

Solve: Time for kicked ball to go Delta x distance

• $$\Delta{x} = (V_{0-x})t$$
• $$\Delta{x} = (\cos{\alpha} * V_0)t$$
• $$t = \frac{\Delta{x}}{(\cos{\alpha}V_0)}$$

Solve: Height of hill - 1st way

• $$\frac{-y_{hill}}{\Delta{x}} = \tan\beta$$
• $$y_{hill} = -(\Delta{x})\tan\beta$$

Solve: Height of hill - 2nd way

• $$y_{hill} = (V_{0-y})t + \frac{1}{2}{g}{t^2}$$
• $$y_{hill} = (\sin{\alpha} * V_0)t + \frac{1}{2}{g}{t^2}$$
• $$y_{hill} = (\sin\alpha * V_0)\left(\frac{\Delta{x}}{(\cos{\alpha}V_0)}\right) + \frac{1}{2}{g}\left(\frac{\Delta{x}}{(\cos\alpha V_0)^2}\right)$$
• $$y_{hill} = \left(\frac{\sin\alpha}{\cos\alpha}\right)\Delta{x} - \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right)$$
• $$\left(\frac{\sin\alpha}{\cos\alpha}\right) = \left(\frac{opposite}{hypotenuse}\right)\left(\frac{hypotenuse}{adjacent}\right) = \tan\alpha$$
• $$y_{hill} = \tan\alpha\Delta{x} - \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right)$$

Solve the two height of hill equations together

• $$y_{hill} = y_{hill}$$
• $$-(\Delta{x})\tan\beta = \tan\alpha\Delta{x} - \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right)$$
• $$-(\Delta{x})\tan\beta = \tan\alpha\Delta{x} - \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right)$$
• $$\tan\alpha\Delta{x} + (\Delta{x})\tan\beta = \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right)$$
• $$\Delta{x}(\tan\alpha + \tan\beta) = \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right)$$
• $$(\tan\alpha + \tan\beta) = \left(\frac{g}{2}\right)\left(\frac{\Delta{x}}{(V_0\cos\alpha)^2}\right)$$
• $$\Delta{x} = \left(\frac{2(\tan\alpha + \tan\beta)(V_0\cos\alpha)^2}{g}\right)$$