Question

My children, I created some notes for projectile motion in Modern English.

Answer 1

\[x(t) = vt\cos\theta \]\[y(t) = vt \sin\theta - \dfrac{1}{2}gt^2\]For the range, we want to find the time that it takes to reach \(0\) height. One instance if this is when we launch the body, and the other is at the end of the motion. Equating the \(y\) component to \(0\),\[0 = vt \sin\theta - \dfrac{1}{2}gt^2 = t\left(v \sin\theta - \dfrac{1}{2}gt\right)\]One solution is \(t = 0\), and the other is \(\dfrac{2v \sin\theta}{g}\). So the time it takes to complete the whole motion is \(\boxed{\dfrac{2v\sin\theta}{g}}\). We can find the range using the \(x\)-component's equation.\[x(t) = v\dfrac{2v\sin\theta}{g} \cos\theta = \boxed{\dfrac{v^2 \sin(2\theta)}{g}}\]Due to reasons of symmetry, the time taken to reach the maximum height is \(\dfrac{v\sin\theta}{g}\). Plugging that in in the \(y\)-equation,\[y_h = v\dfrac{v\sin\theta}{g}\sin\theta - \dfrac{1}{2}g\dfrac{v^2\sin^2\theta}{g^2} = \dfrac{v^2\sin^2\theta}{g} - \dfrac{1}{2}\cdot\dfrac{v^2\sin^2\theta}{g} = \boxed{\dfrac{v^2\sin^2\theta}{2g}} \]Here, we get the maximum height in the projectile.