Question

An object is dropped from a 15 m ledge. How fast it is moving just before it hits the ground?

Answer 1

By v^2 = u^2 + 2gh
v^2 = 0 + 2 x 9.8 x 15
v = √294
v = 17.15 m/s

Answer 2

`v_f = 17.15 m/s`

Step-by-step explanation:

Since the ball is dropped from the height

h = 15 m

so here initial speed of the ball is

`u = 0`

acceleration of the ball due to gravity is given as

`a = 9.8 m/s^2`

now the final speed of the ball is given as

`v_f^2 - v_i^2 = 2 a d`

`v_f^2 - 0 = 2(9.8)(15)`

`v_f = 17.15 m/s`