A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored, so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?

Answer: velocity = 79.6m/s, distance covered = 139.6m and velocity when distance is 10.0m = 61.611m/s

Explanation:

First step is to take a direction of motion, so to avoid issues with negative sign, I take my direction of motion as the positive y axis which makes the parameters positive.

The object is of a free fall thus it has a constant acceleration of g = 9.8m/s².

a) initial velocity (u) = 60m/s, g = 9.8m/s², time taken (t) = 2s, we need to get velocity at that time using the formulae below

v = u + at

v = 60 + (9.8) * 2

v = 60 + 19.6

v = 79.6m/s

b) to get the distance traveled at t = 2s, we use the formulae below

s = ut + 1/2at² where s = distance covered.

s = 60 * 2 + 1/2 * 9.8 * 2²

s = 120 + 1/2 * 9.8 * 4

s = 120 + 9.8 * 2

s = 120 + 19.6

s = 139.6m

c) to get the velocity (v) when distance traveled (s) = 10m, we use the formulae below.

v² = u² + 2as

v² = 60² + 2 (9.8) (10)

v² = 3600 + 196

v² = 3796

v = √3796

v = 61.611m