A ball is thrown directly upward with an initial velocity of 15 m/s from the rooftop of a building that is 30 meters in height. It is caught by a student whose hand is extended from a window. If the students hand is located 18 meters below the rooftop determine the balls velocity when it reaches its maximum height in m/s
To find the velocity of the ball when it reaches its maximum height, we can use the kinematic equation:
vf^2 = vi^2 + 2aΔx
where:
vf = final velocity (when the ball reaches its maximum height)
vi = initial velocity (15 m/s)
a = acceleration due to gravity (-9.8 m/s^2, since the ball is moving upward against gravity)
Δx = change in position (the height the ball needs to reach, 30 meters)
Substituting the given values into the equation, we have:
vf^2 = (15 m/s)^2 + 2(-9.8 m/s^2)(30 m)
vf^2 = 225 m^2/s^2 - 588 m^2/s^2
vf^2 = -363 m^2/s^2
(Note: The negative sign indicates that the velocity is directed upward, opposite to the direction of gravity)
To solve for vf, we take the square root of both sides:
vf = √(-363 m^2/s^2)
Since the velocity cannot have an imaginary value, we can conclude that the ball does not reach its maximum height.