A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What far vertically will it travel before hiting the ground A. 40 m B. 30 m C. 60 m D. 50 m

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

cant without the picture! ?