Applications: Projectile Motion

Core Concepts

Applies parametric equations to model the path of a projectile. This stack covers the standard equations for projectile motion and how to interpret them.

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Applications: Projectile Motion

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In the context of projectile motion, what physical quantities do the parametric functions \(x(t)\) and \(y(t)\) represent?
\(x(t)\) represents the horizontal distance of the object from its starting point at time \(t\). \(y(t)\) represents the vertical height of the object at time \(t\).
What are the standard parametric equations for projectile motion, ignoring air resistance?
For an object launched from height \(h_0\) with initial velocity \(v_0\) at an angle \(θ\) to the horizontal: \(x(t) = (v_0 \cos θ)t\) and \(y(t) = -\frac{1}{2}gt^2 + (v_0 \sin θ)t + h_0\), where \(g\) is acceleration due to gravity.
What is the value of \(g\) (acceleration due to gravity) in the imperial and metric systems?
In the imperial system (feet), \(g \approx 32\) ft/s². In the metric system (meters), \(g \approx 9.8\) m/s².
In the projectile motion equation \(x(t) = (v_0 \cos θ)t\), what does the term \(v_0 \cos θ\) represent?
It represents the initial horizontal component of the velocity, \(v_x\). Since there is no horizontal acceleration (ignoring air resistance), the horizontal velocity is constant.
In the projectile motion equation \(y(t) = -\frac{1}{2}gt^2 + (v_0 \sin θ)t + h_0\), what does the term \(v_0 \sin θ\) represent?
It represents the initial vertical component of the velocity, \(v_y\). The vertical motion is affected by gravity, which is why the equation is quadratic in \(t\).
How can you find the maximum height of a projectile using its parametric equations?
The maximum height occurs when the vertical velocity is zero. Find the time \(t\) when \(y'(t) = 0\). Then, substitute this value of \(t\) back into the \(y(t)\) equation to find the maximum height.
How can you find the total horizontal distance (range) traveled by a projectile launched from the ground?
1. Set the height equation \(y(t) = 0\) and solve for time \(t > 0\). This is the time of flight. 2. Substitute this time \(t\) into the horizontal distance equation \(x(t)\) to find the range.
A cannonball is fired from the ground (\(h_0=0\)) with an initial velocity of \(500\) ft/s at an angle of \(30°\). Write the parametric equations for its motion. Use \(g=32\).
\(x(t) = (500 \cos 30°)t = 250\sqrt{3}t\) \(y(t) = -16t^2 + (500 \sin 30°)t = -16t^2 + 250t\)
If a projectile is launched horizontally from a height \(h_0\) with initial velocity \(v_0\), what is the launch angle \(θ\)?
The launch angle is \(θ = 0°\). This simplifies the equations, as \(\cos(0)=1\) and \(\sin(0)=0\).
For a projectile launched horizontally with velocity \(v_0\) from height \(h\), what are the simplified parametric equations?
Since \(θ=0\), \(\cos(0)=1\) and \(\sin(0)=0\). The equations become: \(x(t) = v_0 t\) \(y(t) = -\frac{1}{2}gt^2 + h\)
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