Question

A 3.96 kg rock is dropped off a 126.3 m tall bridge. How much time, in s, passes before the rock hits the water?

Answer 1

`5.1` s

Explanation:

`x_f = x_i + v_it+ 1/2g t^2`

• Final height `x_f = 0 "m"`

• Initial height `x_i = 126.3 "m"`

• Initial velocity `v_i = 0 "m"/"s"`

• Gravitational acceleration `g = -9.8 "m"/"s"^2`

`=> 0 = (126.3) + 0 + 1/2 (-9.8) t^2`

`=> -126.3 = -4.9 t^2`

`=> t = sqrt(126.3/4.9) = 5.1` s