An object is thrown straight up with an initial velocity of 10 m/s, and there is an air resistance force causing an acceleration of 3 m/s2 opposite the direction of motion. With what speed does the object return to the ground?

Vf = 7.29 m/s

Explanation:

Two force act on the object:

1) Gravity

2) Air resistance

Upward motion:

Initial velocity = Vi = 10 m/s

Final velocity = Vf = 0 m/s

Gravity acting downward = g = - 9.8 m/s²

Air resistance acting downward = a₁ = - 3 m/s²

Net acceleration = a = - (g + a₁) = - (9.8 + 3) = - 12.8 m/s²

(Acceleration is consider negative if it is in opposite direction of velocity)

Now

2as = Vf² - Vi²

⇒ 2 * (-12.8) * s = 0 - 10²

⇒-25.6 * s = - 100

⇒ s = 100 / 25.6

⇒ s = 3.9 m

Downward motion:

Vi = 0 m/s

s = 3.9 m

Gravity acting downward = g = 9.8 m/s²

Air resistance acting upward = a₁ = - 3 m/s²

Net acceleration = a = g - a₁ = 9.8 - 3 = 6.8 m/s²

Now

2as = Vf² - Vi²

⇒ 2 * 6.8 * 3.9 = Vf² - 0

⇒ Vf² = 53. 125

⇒ Vf = 7.29 m/s