A projectile is shot from the edge of a cliff 140 m above ground with an initial speed of 120 m/s at an angle of 38 degrees above the horizontal. What is the time taken by the projectile to hit the ground 140 m below the cliff? (g = 9.8 m/s²)

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.