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 time required to reach the maximum height in seconds
To find the time required to reach the maximum height, we can use the equation:
vf = vi + gt
where vf is the final velocity (which is 0 m/s at the maximum height), vi is the initial velocity (15 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
At the maximum height, the ball is momentarily at rest before starting to fall back down. Therefore, the final velocity is 0 m/s.
0 = 15 + (-9.8)t
Rearranging the equation, we get:
-9.8t = -15
Dividing both sides by -9.8, we have:
t = -15 / -9.8
t ≈ 1.53 seconds (rounded to two decimal places)
So, it takes approximately 1.53 seconds to reach the maximum height.