How do you solve acceleration of gravity formula?

A ball is thrown up at 12m/s. Assume no air resistance and you have to assume that the gravitational field is constant at 9.8ms^-2 then

(for this x will represent the number of seconds since the object was first at a height of 0m)

to work out the speed at each point in time we use the equation 12-9.8x

the 12 is the original speed and the -9.8 is the acceloration towards the earth and the x is the seconds if we calculate the units

12m/s+(-9.8m/s/s*xs) = 12m/s + (-9.8x)m/s this means that the end units are m/s which is speed.

to work out the position we have to times by time again to get m

so the (position at time x) = 12x+(-9.8)*(x^2)

then to find the highest point use the formula x=-b/2a which in this case as we are using the quadratic -9.8x^2+12x+0

-b = -12

2a = -19.6

-12/19.6 = about 0.61 seconds (this is at what time it reaches its highest point) as we have x we can work out the height which is

-9.8(0.61)^2 +12(0.61)+0 = 3.67m

so in answer to the question proposed as an answer for the original question the object would reach 3.67 meters

I think you need to try and clear up what you actually want as an answer for the original question before you can get a proper answer to it.

(james.space.ict@hotmail.co.uk) if you have more questions