# A ball falls from a height

A ball falls from a height h = 100 m. Please find the average speed of its movement in the second half of the way. The initial velocity of the ball is zero and the gravitational acceleration is g = 9.8 m/s^2. The ball may be regarded as a point particle and air resistance neglected.

###### Solution. Posted on Wednesday, April 15, 2015.

We will use the following formula
Yf=Yi+Vi*t-g*t^2/2, where Yf – the final position of the ball, Yi – the initial position of the ball, Vi – the initial velocity, g – gravitational acceleration (9.8 m/s^2) and t – the time which the ball needs to move from the initial position to the final position.

Yf = 0, Yi = h, Vi = 0
0=h-g*t^2/2

So, an equation of time will be
t=√(2h/g)

Now, let’s find an equation of time t1 for the first half of the way
h/2=h-g*t1^2/2

from this,
t1=√(h/g)

The time of the second half of the way will be
t-t1=√(2h/g)-√(h/g)=(√2−1)*√(h/g)

Finally, the average speed will be
V=0.5*h/[(√2−1)*√(h/g)]

V=0.5*100/[(√2−1)*√(100/9.8)] = 37.79 m/s