Aball is dropped from rest at height of 50m above the ground (a) what is its speed just before it hits the ground (b) how long does it take to rech the ground.

To start this question, let's gather some information from the problem. We know that the ball starts from 50 m above the ground, and we're pretty much going to end this problem when it hits the ground level, at 0 m. We're also dropping this ball from rest, meaning that the only acceleration it's going to get is from gravity pulling it towards the earth at -9.8 m/s². This also means that the ball's starting velocity is 0 m/s. With that information, we can make a chart and start the problem!

xi= 50 m
xf= 0 m
vi= 0 m/s
vf= ?
a= -9.8 m/s²
t= ?

Our first step is to find the time. This means we have to find a motion equation that uses the values we know in order to find the time. The first one I can think of is xf = xi + vi*t + 1/2*a*t². Now we can plug in our known values and solve for t.

0 = 50 + 0*t + 1/2*-9.8*t². 
0 = 50 + 0 + -4.9*t². 
-50 = -4.9*t². 
10.204 = t²
3.194 = t

So our value for time should be around 3.194 seconds. Now that we know this, we can find another motion equation that will let us solve for vf using our known values. One that fits what we need is vf = vi + a*t. Using this, we can plug in and solve for the vf.

vf = 0 + -9.8*3.194
vf = -31.3012

This means our final velocity, or the speed the ball is going just before it makes contact with the ground is -31.3012 m/s.

The reason this value is negative is because of the way we set the equation up. It is possible to set up the equation in such a way that everything comes out to the same values, but some things would switch signs. For example, if we set up the equation so that our starting height was -50 m and our end height was 0 m, we would've had a positive value for gravity, and thus our final velocity would have been positive. However, both ways of setting up this problem are correct and will get you the right answer!