Question

A ball with mass m kg is thrown upward with initial velocity 12 m/s from the roof of a building 15 m high. Neglect air resistance. Use g = 9.8 m/s. Find the maximum height above the ground that the ball reaches.

Answer 1

The maximum height above ground reached by the ball is 22.35 meters, calculated using kinematic equations.

Step-by-step explanation:

Finding the Maximum Height Above Ground of a Thrown Ball

To find the maximum height above the ground that a ball reaches when thrown upward, you can use the principles of kinematics in physics. The ball is thrown upward with an initial velocity of 12 m/s from a building 15 m high. Using the formula for maximum height (h) when initial velocity (u), acceleration due to gravity (g), and the height of the building (h0) are known:

h = h0 + (u^2) / (2g)

First, convert the initial velocity to meters per second (if it's not already), then plug in the values:

h0 = 15 m (height of building)

u = 12 m/s (initial velocity)

g = 9.8 m/s2 (acceleration due to gravity)

Now, calculate the additional height the ball reaches above the building:

additional height = (u^2) / (2g) = (12 m/s)^2 / (2 * 9.8 m/s2) = 144 / 19.6 = 7.35 m

Then, add the height of the building to find the total maximum height above ground:

maximum height = h0 + additional height = 15 m + 7.35 m = 22.35 m

Therefore, the maximum height above the ground the ball reaches is 22.35 meters.