Answer:
17 seconds
Explanation:
In the y direction:
y = y₀ + v₀ᵧ t + ½ gt²
0 = 140 + (120 sin 38) t + ½ (-9.8) t²
4.9 t² - 73.9 t - 140 = 0
Solve with quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 73.9 ± √((-73.9)² - 4(4.9)(-140)) ] / 9.8
t = -1.7, 16.8
Since t can't be negative, t = 16.8. Rounding to 2 sig-figs, the projectile lands after 17 seconds.
