(a) 70.1 m
The height of the ball at time t is described by the following equation:
[tex]h(t) = h_0 - \frac{1}{2}gt^2[/tex] (1)
where
[tex]h_0 = 75 m[/tex] is the initial height of the ball
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
t is the time
Since we want to know the heigth of the ball after 1 second, we just need to substitutite t=1 s into the formula, and we find:
[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
To find the time it takes for the ball to reach the ground, we have to find the time t at which the height of the ball h(t) is zero: h(t) =0. Using eq.(1), this means:
[tex]0=h_0 - \frac{1}{2}gt^2[/tex]
And re-arranging we find:
[tex]t=\sqrt{\frac{2h_0}{g}}=\sqrt{\frac{2(75 m)}{9.8 m/s^2}}=3.9 s[/tex]
