Question

A projectile is launched from ground level with an initial velocity of 64 feet per second and at an angle of 45 degrees with the horizontal. How i determine the parametric equations that model the projectiles path, find the time it takes to hit the ground the horizontal distance it has traveled when it hits the ground and the maximum height reached by the projectile to the nearest foot?

Answer 1

Start with the following equations:\[x_f=x_0+v_0xt+\frac{1}{2}ayt^2\space and \space y_f=y_0+v_0yt+\frac{1}{2}ayt^2\]

Answer 2

Is there a way to do this using the parametric equations...
\[x =(v _{0}\cos \theta)t\] and \[y=-16t ^{2}+(v _{0}\sin \theta)t+s _{0}\]

Answer 3

Yes, you will need to break the velocity into its components which give you \[v _{0x}=v _{0}\cos \theta \space and \space v _{0y}=v _{0}\sin \theta\]

Answer 4

Can you help me get this problem started at least, I have tried and I am not sure how to do it. Plz?

Answer 5

Okay let's first look at what your known and unknown are:
Known:
\[v_0=64\frac{f}{s}, \space \theta=45, \space x_0=0, \space y_0=0, \space t_0=0\]
Unknown:\[x_f, \space t_f,\space y_f\]

Answer 6

So let's find the time it takes to go up.
We need to break the initial velocity into its components so:\[v_{0x}=v_0\cos 45 \rightarrow 64\frac{f}{s} \times \cos 45=45.25\frac{f}{s}\]\[v_{0y}=v_{y}\sin45 \rightarrow 64\frac{f}{s} \times \sin45=45.25\frac{f}{s}\]
\[y_f=y_0+v_0yt+\frac{1}{2}ayt^2 \rightarrow 0+v_{0y}t-\frac{1}{2}g t^2\] subtract\[\frac{1}{2}g t^2=v_{0y}t \rightarrow \frac{1}{2}g t=v_{0y} \rightarrow t=\frac{2v_{0y}}{g}\]

Answer 7

So for Y we use the equation \[y=\frac{v_{fy}^2-v_{iy}^2}{2g} \rightarrow \frac{45.25\frac{f}{s}^2 - 0\frac{f}{s}^2}{2\times 32.15\frac{f}{s}}=31.84feet\]

Answer 8

So x is just a simplified version as most things are zero\[x=v_{x0}t \rightarrow 45.25\frac{f}{s} \times 2.81s=127.38feet\]