A ball is thrown horizontally from the top of a building which is 30.6 meters tall. It lands 50.0 meters from the base of the building.

What was the launch speed of the ball?

First, we need the time before the ball hits the ground. This would be:

0=30.6-4.9t²
4.9t²=30.6
t=2.5 secs
50 meters/2.5 secs=20 m/s as the speed of the ball
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To find the launch speed of the ball, we can use the kinematic equation for horizontal motion:

d = v*t

Where:
d = distance traveled (50.0 meters)
v = launch speed (unknown)
t = time of flight (unknown)

Since the ball is thrown horizontally, the vertical motion does not affect the horizontal distance traveled. Hence, we can ignore the height of the building for now.

Now let's find the time of flight. We can use the equation for vertical motion:

h = 1/2 * g * t^2

Where:
h = height of the building (30.6 meters)
g = acceleration due to gravity (9.8 m/s^2)
t = time of flight

Rearranging the equation to solve for t:

t = √(2h / g)

Substituting the values h = 30.6 meters and g = 9.8 m/s^2:

t = √(2 * 30.6 / 9.8)
t ≈ 2.04 seconds

Now that we know the time of flight, we can use the horizontal motion equation to find the launch speed:

d = v * t

Substituting the values d = 50.0 meters and t = 2.04 seconds:

50.0 = v * 2.04

Solving for v:

v ≈ 50.0 / 2.04
v ≈ 24.5 m/s

Therefore, the launch speed of the ball was approximately 24.5 m/s.