Answer:
First, let's think in the vertical problem:
The acceleration will be the gravitational acceleration:
g = 9.8 m/s^2
a = -9.8 m/s^2
For the velocity, we integrate over time:
v(t) = (-9.8 m/s^2)*t + v0
Where v0 is the initial velocity, in this case v0 = 30m/s.
v(t) = (-9.8 m/s^2)*t + 30m/s
Now, for the position we integrate again over time, and get:
P(t) = (1/2)*(-9.8 m/s^2)*t^2 + 30m/s*t + p0
Where p0 is the initial position, as the ball is launched from the ground, we can use p0 = 0m
p(t) = (-4.9m/s^2)*t^2 + 30m/s*t
Now, the maximum vertical height is reached when:
v(t) = 0m/s = -9.8m/s^2*t + 30m/s
t = 30m/s/9.8m/s^2 = 3.06s
Now we can evaluate the vertical position in t = 3.06s
p(3.06s) = (-4.9m/s^2)*(3.06)^2 + 30m/s*3.06 = 62m
So, rounding down, the correct option is: C. 60 m
