Verified answer

Yes here is a situation where you can use this eqn. --

" If a body is dropped from a height of ' h ' m above the ground., calculate the height h if it takes ' t ' sec. for the body to reach the ground. Take accn. due to gravity = (g) m.s⁻² "

Look how it is used---

Since the body is dropped , its initial velocity ( say u ) = 0 m/s. Applying Newton's Law of motion we get --

=> h = u t + ½gt²

But u = 0 in this case . Hence ----

=> h = 0 (t) + ½gt²

=> h = ½gt² ........................... Your Eqn.

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

.

Your equation is a special case of:

x_f = x_i + v_ix∙t + ½a∙t²

So, you can use it when:

→ The acceleration "g" is constant, because the general equation above is derived from the definition of acceleration, a = dv/dt, under the assumption that the acceleration is constant. In the context of vertical motion near the Earth, this means it won't work for heights of hundreds of kilometers.

→ v_ix = 0. In the context of vertical motion, that means the object must be dropped, rather than thrown.

→ x_i = 0, or h represents x_f - x_i. Setting x_i = 0 should never be a problem, as you can always choose a coordinate system that makes it true.

And yes, you can use it for other constant acceleration problems because, as noted above, h = ½gt², is just a special case of a more general constant acceleration kinematics formula. For that reason, I would discourage you from remembering h = ½gt². You'll probably need to remember the general formula anyway to solve other problems, and if you use it and substitute the values of zero when appropriate, you'll get the same result as using h = ½gt², but there will be one less equation for you to remember. Also, in my experience, when a student memorizes a very special case like h = ½gt², they'll often forget the circumstances under which it is valid (for example, when the object is dropped and not thrown), and use it when it doesn't apply. This isn't an issue if you always start with the general formula.

Here is a question that this would work for:

A weight is dropped from a height h. Assume it is dropped from rest, not thrown down. The weight falls for exactly t seconds before it hits the ground. Ignoring air resistance from what height h was the weight dropped.

If you want to use a different acceleration than gravity, the equation works, but the physical situation would be different from what I suggested. For example, If you assumed a rocket accelerated at a constant acceleration of a, straight up, the height after time t would be 1/2at^2

That is the "short form" of one of Newton's Equations of motion.

The Long Form is:

Displacement = 0.5 * acceleration * time^2 + Original velocity * time + Original Displacement

D = 0.5At^2 + Vo*t + Do

If your Original Displacement is 0, that term drops out

If your Original Velocity is 0 (object starts from rest), that term drops out

Leaving

D = 0.5At^2

Works for any acceleration including gravity.

If you have any 2 of the 3 variables, you can solve for the 3rd.