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???
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".
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)
To find the height or time, you need to use the appropriate formula based on the situation. It seems like you are referring to the equation for the height of an object in free fall. In this case, the correct formula is h = (1/2) * g * t^2, where h represents the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
To find the time, you can rearrange the formula to solve for t. Follow these steps:
1. Start with the formula h = (1/2) * g * t^2.
2. Divide both sides of the equation by (1/2) to isolate the t^2 term: h / (1/2) = g * t^2.
3. Simplify the left side of the equation: 2h = g * t^2.
4. Divide both sides of the equation by g: 2h / g = t^2.
5. Take the square root of both sides to solve for t: sqrt(2h / g) = t.
Therefore, the formula to find time is t = sqrt(2h / g).
So, to find the time, you can plug in the values of h and g into the formula t = sqrt(2h / g), and then evaluate it using a calculator if necessary.