a ball is thrown straight up with an initial velocity of 25ft per second from the top of a building 200 feet tall. When does the ball pass pass the roof on its

Mike C. | Enthusiastic Tutor for Middle and High School StudentsEnthusiastic Tutor for Middle and High S...

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Remember conservation of energy? It says that the total energy in a closed system will always remain constant. When it comes to free-fall, we can go one further and say that the kinetic energy of the system at one height level is the same, regardless of direction of motion (for convenience, we'll ignore air resistance). In other words, at the same height (here, the level of the roof), the speed of the ball will be the same in the system, so since the ball is thrown upwards at 25ft/s, it will also come down at 25 ft/s (25 ft/s is 7.62 m/s, by the way, since we'll need to convert to metric measurements).

Because the ball left your hand at 25ft/s (7.62m/s) upwards and came back down at the same speed, this implies that there was an overall velocity change of 50ft/s (15.24m/s). The gravitational pull of the Earth (9.8m/s^{2}) did all of the work in changing that velocity, so we can calculate the total time by simply dividing the velocity change by the acceleration.

The answer, therefore, is (15.24m/s)/(9.8m/s^{2}) or about 1.555 seconds.

Austin P. | Astrophysics Major for Math, Science, Writing, or Music TutoringAstrophysics Major for Math, Science, Wr...

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Your answer got cut off, but I'm going to guess you were asked at what time will the ball pass by the roof on it's way down. For this question you don't need a quadratic equation. All you need is the kinematic equation V_{f}=V_{i}+at, where V_{f} is our final velocity, V_{i} is our initial velocity, a is our acceleration (in this case gravity), and t is time. We also need to know the fact that the ball will take the same time going up as it will coming down since acceleration due to gravity is constant.

First, we convert our given initial velocity to meters/second; 25 ft/s=7.5m/s. Accel. due to gravity will be 9.8 m/s^{2}. Plugging these values into our kinematic equation to find the time it takes for the ball to reach it's highest point and then stop, we get t=0.765 seconds. Doubling this to add the time it takes to come back down we get T_{f}=1.53 seconds, our final answer.