A student is measuring the acceleration due to gravity, g. They drop a piece of card from rest, from a vertical height of 0.75m above a light gate. The light gate measures the card's speed as it passes to be 3.84 m/s. Calculate an estimate for g.
This question is simply a matter of finding and applying the correct equation of motion.
First, draw a diagram - even if the situation in the questions seems really simple, it's always useful to draw a diagram.
The card is being dropped from rest; as the only force acting on it is gravity, the acceleration - and therefore final velocity - will be purely vertical. The question also specifies that the card is dropped from a given vertical distance - this is all good news as we will not have to resolve anything into horizontal and vertical components.
Now we write down all the quantities that we might find in an equation of motion that relate to the question.
s = distance travelled = 0.75m
u = initial velocity = 0 m/s
v = final velocity = 3.84 m/s
a = acceleration = g, to be found
t = time elapsed. We are not given t, nor are we at any point required to find it - so as far as we are concerned t is irrelevant!
The only quantities that matter are s, u, v and a; so we just find the equation of motion that includes these four, and substitute our values in.
From either memory or the formula sheet, we have the equation:
v2 - u2 = 2as.
Simply rearrange in terms of the value we are trying to find, in this case a, and plug in the numbers to get our estimate for g,
a = 9.83 m/s2.
You might notice this is actually slightly different from the accepted value of 9.81 - this is fine! The exam board will rarely have you calculate a known constant that comes out exactly right; this is to prevent you just looking up the answer in the formula booklet. As our answer is very close, we can be confident that the calculations are correct.
