# Parametric equation projectile problem

I'm trying to find the distance that the bullet hits the ground with these equations:

$$x = (v\cos(a))t\quad y = (v\sin(a))t-\frac12gt^2\\[8pt] V = 500\ \text{m/s}\ \text{and}\ a = \frac\pi6, g = 9.8$$

I assumed that $y = 0$ and plugged in all values but then I got stuck when I tried to solve it.

\begin{align} 0 &= 250t-\frac129.8t^2\\ \implies -t^2+t &= 0 \end{align}

That doesn't seem right to me at all.

I'm also having trouble with the second part of the equation, What will be the maximum height reached by the bullet?

I'd appreciate some help on solving the first and setting up the second, I'm stuck.

The full problem:

If a projectile is fired with an initial velocity of $v_0$ meters per second at an angle $\alpha$ above the horizontal and air resistance is assumed to be negligible, then its position after $t$ seconds is given by the parametric equations $x = (v_0\cos\alpha)t$ and $y = (v_0\sin\alpha)t-\frac12gt^2$ where $g$ is the acceleration due to gravity $\left(9.8\frac m{s^2}\right)$.

a) If the gun is fired with $\alpha=\frac\pi6$ and $v_0=500\frac ms$, when will the bullet hit the ground? What will be the maximum height reached by the bullet?

b) Show that the path is parabolic by eliminating the parameter.

Hints:

1. The expression for $y$ is quadratic in $t$. $t=0$ is one solution when $y=0$. What is the other?
2. Taking the time derivative with respect to $y$ gives the $y$ velocity. What does it equal at the maximum?

$0=(250)t−(1/2)9.8t^2$

So far so good.

factor:

$t(4.9 t - 250) = 0$

$t = 0, t = \frac {250}{4.9}\approx 51$

The projectile is at ground level when it is launched, and hits ground again at $t = 51$

As for the apex. The projectile traces out a parabolic path, and parabola are symmetric, with the vertex on the line of symmetry. i.e. the line of symmetry is halfway between the zeros. Or, it is at $t = \frac {250}{9.8}$

• Can't believe I couldn't see that it could be factored. I ended up doing a quadratic and ended up with the same so I guess it's all good. Appreciate the tips. – cout Mar 13 '17 at 16:57
• I frequently see people use calculus when elementary geometry or algebra will suffice. Not that it is wrong, but it might be more complicated than necessary. People seem to like to use the tools they have recently acquired. – Doug M Mar 13 '17 at 17:02