Question

A student drops a rock from a bridge to the

water 12 m below.
With what speed does the rock strike
the water?
Answer in units of m/s ...?

Answer 1

, let us use the expression v = ./2gS in both the cases.
a) In the first case g = 9.8 m/s^2 and S = 12 m.
So v = 14.21 m/s

Answer 2

15.49m/s

Step-by-step explanation:

Using one of the equations of motion;

`v^(2)` =
`u^(2)` + 2gs

Where;

v = the final velocity of the rock.

=> This is the speed at which the rock strikes the water.

u = initial velocity of the rock.

=> The rock is just dropped from that height. That means the initial velocity is zero (0).

=> u = 0

g = the acceleration due to gravity.

=> Since the rock moves downwards, the acceleration due to gravity is positive and let it have a value of 10m/
`s^(2)`

=> g = 10m/
`s^(2)`

s = the distance covered by the rock

=> s = 12m

Substituting these values into the equation above gives

`v^(2)` =
`0^(2)` + 2(10 x 12)

`v^(2)` = 240

v =
`√(240)`

v = 15.49m/s

Therefore the velocity(speed) with the rock strikes the water is 15.49m/s