Can somebody please help me with this physical science problem

Question

a child dropped a marble from a bridge 120 m high. How long did it take for the marble to reach the water. and what was the velocity of the marble before it hit the water.

please help me im really confussed

Answer 1

A simple equation here should be all you need.  I'm not sure what your teacher has given you, but I would use x = x0 + v0 * t + (1/2) * a * t^2. (t^2 means t squared, or t to the second power.  x0 means initial.  x is position, v is velocity, a is acceleration, and t is time.)

v0 is 0, and our x is zero.  So we have

0 = x0 + a * t^2

Now, our acceleration near the surface of the earth is around 9.8 m/s^2 downward.

Solve for t:

t = square root of (-x0 / a)

Plug in your numbers: square root of (- 120 m / (- 9.8 m/s^2)) gives me around 3.5 seconds.

Answer 2

First we'll find velocity by the equation v^2=u^2 +2as

(v is final velocity in m/s, u is initial velocity in m/s, a is acceleration in m/s^2, s is distance-not to be confused with seconds- in m)

v^2= 0^2 + 2(9.8m/s^2)(120m)

v^2= 2352 m^2/s^2

v= 48.5 m/s

From there we can find the time it took to hit the water with the equation v=u+at

(where v is final velocity in m/s, u is initial velocity in m/s, a is acceleration in m/s^2, and t is time in s)

48.5m/s = 0 + (9.8m/s^2)t

4.95 s = t

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p.s. 9.8 is the acceleration due to gravity