Bullet Trajectory

Question

If a bullet is shot at a 5% angle with a speed of 76.2 m/s and a starting height of 1.44 m how far will the bullet travel before hitting the ground? 

Answer 1

sin5= about 0.087

0.087(76.2) = the upward initial velocity

h(t) = -4.9t^2 +0.087(76.2)t + 1.44

set h(t) = 0 and solve for t

-4.9t^2+ 6.63t + 1.44 = 0

use the quadratic formula

t = 6.63/9.8 +(1/9.8)sqr(6.63^2 +4(4.9)(1.44))

=about 0.68+ 0.87

= 1.65 seconds until it hits the ground

plug that t value into a formula for distance