You are going to have to describe the situation. Where is the height measured from? Is something falling beginning at time t=0? Does g = 9.8 m/s^2?
If an object is dropped at time t=0 in a gravitational field where g = 9.8 m/s^2, then the distance it falls in time t is
h = (g/2)*t^2 = 4.9 t^2
That looks like "something else comletely".
how do you find the height or time?
is it h=squreroot4.9/t
or h=squareroott/4.9 or is it something else complettly???
2 answers
Well, we seem to be playing "Here is a choice of answers. What is the question?"
Now if I drop a rock in frictionless air on earth, the acceleration is about 9.8 m/s^2 down
In that case the speed is about
v = 9.8 t
and the distance is about
h =(1/2)(9.8) t^2
or
h= 4.9 t^2
That means if you know h (the height) and want to know how long it took to fall, t
t = sqrt (h/4.9)
From that you can get the speed when it reached the ground at h below the dropping point
v = 9.8 t
v = 9.8 sqrt (h/4.9)
Now if I drop a rock in frictionless air on earth, the acceleration is about 9.8 m/s^2 down
In that case the speed is about
v = 9.8 t
and the distance is about
h =(1/2)(9.8) t^2
or
h= 4.9 t^2
That means if you know h (the height) and want to know how long it took to fall, t
t = sqrt (h/4.9)
From that you can get the speed when it reached the ground at h below the dropping point
v = 9.8 t
v = 9.8 sqrt (h/4.9)