If a projectile is fired with an initial velocity ofv_{0}meters per second at an angleαabove the horizontal and air resistance is assumed to be negligible, then its position aftertseconds is given by the parametric equations

$\displaystyle x = (v_{0}\cos\theta)t$

$\displaystyle y = (v_{0}(\sin\theta)t =- \dfrac{1}{2}gt^{2}$

wheregis the acceleration due to gravity (9.8 m/s^{2}). (Round your answers to the nearest whole number.) (a) If a gun is fired withα= 30° andv_{0}= 900 m/s.

$\displaystyle x = (900 m/s\cos30^{o})t$

$\displaystyle y = (900 m/s\sin30^{o})t =- \dfrac{1}{2}(9.8 m/s)t^{2}$

$\displaystyle x = (900 m/s\dfrac{\sqrt{3}}{2})t$

$\displaystyle y = (900 m/s\dfrac{1}{2})t - \dfrac{1}{2}(9.8 m/s)t^{2}$

When will the bullet hit the ground?

$\displaystyle (900 \dfrac{1}{2})t - \dfrac{1}{2}(9.8)t^{2} = 0$

$\displaystyle 300t - 4.9t^{2} = 0$

$\displaystyle t(300 - 4.9t) = 0$

$\displaystyle t = 0s$

$\displaystyle t = 61.2s$

How far from the gun will it hit the ground?

$\displaystyle t = \dfrac{300}{4.9} = 61$

What is the maximum height reached by the bullet?

(b) Find the equation of the parabolic path by eliminating the parameter.