A ball is thrown straight up, it rises 12.5 m before it starts falling back down.

a) What was its initial speed?

b) How much time is it in the air?

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a) v^2 = 2gh

v = sqrt(2gh) = sqrt(2*9.8*12.5) = 15.65 m/s

b) v = gt

t = v/g = 15.65/9.8 = 1.60 sec

The problem shall be formulated slightly better. If a ball was thrown up from the ground, the Robert's solution is right. However, the problem does not state this, so this is just an assumption, however logical it may be.

As for the time in the air, it is twice the time to get to the top, that is

t=2*v/g=3.2 s

You can formally write the equation of motion:

y=y_{0}+vt-gt^{2}/2; Here y_{0} is the initial position, assumed to be zero (ground level), v is the initial velocity, pointed up, which is also positive y-direction, therefore + sign in front of vt term, g--acceleration due to gravity, pointing down, hence negative sign in front of the term gt^{2}/2. Once you know the velocity (see Robert's solution for v), you know full equation of motion:

y=15.65t-4.9t^{2}

Ball falls back to the ground, which means that its y-coordinate is zero. Set y=0, you will obtain:

15.65t-4.9t^{2}=0; Factor t out, you will get:^{
}

t*(15.65-4.9t)=0;

There are two solutions to this equation:

t=0 s;

t=15.65/4.9=3.2 s;

First solution is trivial and corresponds to the initial position of the ball, just before it was thrown up into the air.

Second one corresponds to the moment when the ball falls back to the ground. This is the one you need.

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