Parametric equation projectile problem

Question

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.

Answer 1

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?

Answer 2

$0=(250)t−(1/2)9.8t^2$

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 $t = \frac {250}{9.8}$