Question

a ball is thrown straight up into the air with an initial speed of 18.2 m/s. How far has it gone after 1.00s?

Answer 1

s= ut + 0.5 at^2

s= distance
u= initial speed (= +18.2m/s)
a= -9.81m/s^2
t= 1s

Answer 2

Be careful with the signs

Initial speed is in opposite direction to acceleration

Answer 3

i thought a= -9.80? @MrNood

Answer 4

it really doesn't matter in this case ( the difference you are talking about is 0.1%)

I believe -9.81 is a more accurate figure.

However - use whatever value you want - but put it into the equation above and solve for s

Answer 5

i got 66.2 but it said wrong @MrNood

Answer 6

to quote from Wikipedia:

The precise strength of Earth's gravity varies depending on location. The nominal "average" value at the Earth's surface, known as standard gravity is, by definition,[2] 9.80665 m/s2

Answer 7

That's cos it IS wrong.

I have given you the equation and all the values.

If you show your working I will help you see where you made the mistake, but I will not post the answer

Answer 8

start by calculating
\[\frac{at ^{2}}{ 2 } \]

using the values I gave above

Answer 9

gone offline as soon as I said I wouldn't give answer