Question

The position of a projectile fired with an initial velocity $v_0$feet per second and at an angle u to the horizontal at the end of t seconds is given by the parametric equations

$x=\left(v_0\cos \theta \right)ty=\left(v_0\sin \theta \right)t-16t^2$

(a) Obtain the rectangular equation of the trajectory and identify the curve.

(b) Show that the projectile hits the ground $\left(y=0\right)$when$t=\frac{1}{16}{v}_0\sin \theta$

(c) How far has the projectile traveled (horizontally) when it strikes the ground? In other words, find the range R.

(d) Find the time t when x = y. Then find the horizontal distance x and the vertical distance y traveled by the projectile in this time. Then compute $\sqrt{x^2+y^2}$. This is the distance R, the range, that the projectile travels up a plane inclined at 45° to the horizontal $\left(x=y\right)$.

Short answer

(a) The rectangular equation of the trajectory is $y=x\tan \theta -\frac{16x^2}{V_0\cos {\theta }^2}$the curve is a parabola

(b) The projectile hits the ground $\left(y=0\right)$when $t=\frac{1}{16}{v}_0\sin \theta$

(c) The range of the projectile is localid="1648354391781" $R=\frac{v_o^2\sin 2\theta }{32}$

(d) The time t when x = y is $t=\frac{v_0\left(\sin \theta -\cos \theta \right)}{16}$,

the horizontal distance x traveled by the projectile in this time is $x=\frac{{v_0}^2\left(\sin \theta \cos \theta -{\cos }^2\theta \right)}{16}$

the vertical distance y traveled by the projectile in this time is $y=\frac{{v_0}^2\left({\sin }^2\theta -\sin \theta \cos \theta -{\left(\sin \theta -\cos \theta \right)}^2\right)}{16}$

the range R, that the projectile travels up a plane inclined at 45° to the horizontal is

$R=\sqrt{{\left(\frac{{v_0}^2\left(\sin \theta \cos \theta -{\cos }^2\theta \right)}{16}\right)}^2+{\left(\frac{{v_0}^2\left({\sin }^2\theta -\sin \theta \cos \theta -{\left(\sin \theta -\cos \theta \right)}^2\right)}{16}\right)}^2}$

Step-by-step solution

Step 1. Given Information

The position of a projectile fired with an initial velocity $v_0$feet per second and at an angle u to the horizontal at the end of t seconds is given by the parametric equations

role="math" localid="1648352120463" $x=\left(v_0\cos \theta \right)ty=\left(v_0\sin \theta \right)t-16t^2$

Part (a) of Step 1. The rectangular equation of the trajectory 

The given equation $x=\left(v_0\cos \theta \right)ty=\left(v_0\sin \theta \right)t-16t^2$

Let localid="1648353657514" $\left.x=\left(v_0\cos \theta \right)t\text{and}y=v_0\sin \theta \right)t-16t^2$

also, the equation localid="1648354036080" role="math" $x=\left(v_0\cos \theta \right)t$can be written as

localid="1648353676261" $t=\frac{x}{\left(v_0\cos \theta \right)}$

Substitute the value of t in the equation $y=V_0\sin \theta t-16t^2$

localid="1648353689101" $y=\left(v_0\sin \theta \right)\left(\frac{x}{\left(v_0\cos \theta \right)t}\right)-16\left(\frac{x}{\left(v_0\cos \theta \right)t}\right)$

Simplify the above equation by using $\frac{\sin \theta }{\cos \theta }=\tan \theta$

localid="1648353698640" $y=x\tan \theta -\frac{16x^2}{v_0\cos {\theta }^2}$

The obtained equation is of the form $y=a{x}^2+bx+c$where localid="1648353705660" $a=\frac{-16}{v_0\cos {\theta }^2},b=\tan \theta ,c=0$

and the curve is a parabola.

Part (b) of Step 1. The projectile hits the ground when t=116v0sinθ

Substitute the value of t in the equation $y=\left(V_0\sin \theta \right)t-16t^2$and solve

role="math" localid="1648358812734" $$\begin{gathered} y=\left(V_0\sin \theta \right)t-16t^2 \\ y=\left(V_0\sin \theta \right)\left(\frac{1}{16}{v}_0\sin \theta \right)-16{\left(\frac{1}{16}{v}_0\sin \theta \right)}^2 \\ y=0 \end{gathered}$$

Part (c) of Step 1. The range of the projectile 

Substitute value of $t=\frac{1}{16}{v}_0\sin \theta$in the equation $x=\left(v_0\cos \theta \right)t$and solve

role="math" localid="1648354294590" $$\begin{gathered} x=\left(v_0\cos \theta \right)t \\ x=\left(v_0\cos \theta \right)\frac{1}{16}{v}_0\sin \theta \\ x=\frac{v_o^2\sin \theta \cos \theta }{16} \end{gathered}$$

Substituting $$\begin{gathered} \sin 2\theta =2\sin \theta \cos \theta \\ \frac{\sin 2\theta }{2}=\sin \theta \cos \theta \end{gathered}$$in the above equation

role="math" localid="1648354343681" $$\begin{gathered} x=\frac{v_o^2\sin \theta \cos \theta }{16} \\ x=\frac{v_o^2\sin 2\theta }{32} \end{gathered}$$

The range of the projectile is$R=\frac{v_o^2\sin 2\theta }{32}$

Part (d) of Step 1. Time t, Horizontal, and vertical distance

The time, t when $x=y$can be calculated by solving the equation localid="1648358254796" $\left(v_0\cos \theta \right)t=\left(v_0\sin \theta \right)t-16t^2$

localid="1648359396922" $$\begin{gathered} \left(v_0\cos \theta \right)t=\left(v_0\sin \theta \right)t-16t^2 \\ {v}_0\cos \theta =v_0\sin \theta -16t \\ 16t=v_0\sin \theta -v_0\cos \theta \\ t=\frac{v_0\sin \theta -v_0\cos \theta }{16} \\ t=\frac{v_0\left(\sin \theta -\cos \theta \right)}{16} \end{gathered}$$

Horizontal distance x traveled by the projectile in this time can be calculated by substituting the value of t in the equation localid="1648358569949" $x=\left(v_0\cos \theta \right)t$

localid="1648359473306" $$\begin{gathered} x=\left(v_0\cos \theta \right)t \\ x=\left(v_0\cos \theta \right)\left(\frac{v_0\left(\sin \theta -\cos \theta \right)}{16}\right) \\ x=\frac{{v_0}^2\left(\sin \theta \cos \theta -{\cos }^2\theta \right)}{16} \end{gathered}$$

Vertical distance y traveled by the projectile in this time can be calculated by substituting the value of t in the equation $y=\left(V_0\sin \theta \right)t-16t^2$

localid="1648359514315" $$\begin{gathered} y=\left(V_0\sin \theta \right)t-16t^2 \\ y=\left(V_0\sin \theta \right)\left(\frac{v_0\left(\sin \theta -\cos \theta \right)}{16}\right)-16{\left(\frac{v_0\left(\sin \theta -\cos \theta \right)}{16}\right)}^2 \\ y=\frac{{v_0}^2\left({\sin }^2\theta -\sin \theta \cos \theta \right)}{16}-\frac{{\left(v_0\left(\sin \theta -\cos \theta \right)\right)}^2}{16} \\ y=\frac{{v_0}^2\left({\sin }^2\theta -\sin \theta \cos \theta -{\left(\sin \theta -\cos \theta \right)}^2\right)}{16} \end{gathered}$$

The range of the that the projectile travels up a plane inclined at 45° to the horizontal can be calculated by substituting the value of x and y in the equation $\sqrt{x^2+y^2}$

localid="1648359575666" $$\begin{gathered} R=\sqrt{x^2+y^2} \\ R=\sqrt{{\left(\frac{{v_0}^2\left(\sin \theta \cos \theta -{\cos }^2\theta \right)}{16}\right)}^2+{\left(\frac{{v_0}^2\left({\sin }^2\theta -\sin \theta \cos \theta -{\left(\sin \theta -\cos \theta \right)}^2\right)}{16}\right)}^2} \end{gathered}$$