Question

What distance would you predict for the ball after it has fallen for 10.0 seconds?

velocity increases by 10 for every second

constant acceleration of 10

Answer 1

Assuming the ball begins with initial velocity of 0:
\[x_{f}=x_{0}+v_{0}t+\frac{1}{2}at^2\]
Using this formula
\[x_{f}=0+0(10s)+\frac{1}{2}(10\frac{m}{s^2})(10s)^2\]

Can you solve that?