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. What is the balls maximum height above the rooftop in meters
The ball is thrown directly upward with an initial velocity of 15 m/s. As it rises, it will eventually reach a maximum height before falling back down.
The time it takes for the ball to reach the highest point can be determined using the equation:
v = u + at,
where v is the final velocity (0 m/s at the top), u is the initial velocity (15 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
Rearranging the equation to solve for t:
t = (v - u) / a = (0 - 15) / -9.8 = 1.53 seconds (approximated to two decimal places).
Now we can calculate the maximum height. The maximum height can be determined using the equation:
s = ut + (1/2)at^2,
where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration due to gravity.
Substituting the known values:
s = (15 * 1.53) + (0.5 * -9.8 * (1.53)^2) = 23.19 - 11.19 = 12 meters.
Therefore, the ball reaches a maximum height of 12 meters above the rooftop.