a) 70.1 m
The ball is moving by uniformly accelerated motion, with constant acceleration [tex]g=9.8 m/s^2[/tex] (acceleration due to gravity) towards the ground. The height of the ball at time t is given by the equation
[tex]h(t)=h_0 - \frac{1}{2}gt^2[/tex]
where [tex]h_0 = 75 m[/tex] is the height of the ball at time t=0. Substituting t=1 s, we can find the height of the ball 1 seconds after it has been dropped:
[tex]h(1 s)=75 m - \frac{1}{2}(9.8 m/s^2)(1 s)^2=70.1 m[/tex]
b) 3.9 s
We can still use the same equation we used in the previous part of the problem:
[tex]h(t)=h_0 - \frac{1}{2}gt^2[/tex]
This time, we want to find the time t at which the ball hits the ground, which means the time t at which h(t)=0. So we have
[tex]0=h_0 - \frac{1}{2}gt^2[/tex]
And solving for t we find
[tex]t=\sqrt{\frac{2h_0}{g}}=\sqrt{\frac{2(75 m)}{9.8 m/s^2}}=3.9 s[/tex]
