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. Is the ball rising or falling at t=3.5 seconds
To determine whether the ball is rising or falling at t = 3.5 seconds, we need to find the position of the ball at this time.
Let's consider the motion of the ball in two parts: the upward motion and the downward motion.
First, let's calculate the time for the ball to reach its highest point. The initial vertical velocity of the ball is 15 m/s, and the acceleration due to gravity is -9.8 m/s² (negative because it acts in the opposite direction of the upward motion). Using the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time:
0 = 15 - 9.8t_up,
where t_up is the time taken for the ball to reach its highest point. Solving this equation:
9.8t_up = 15,
t_up = 15 / 9.8 ≈ 1.53 seconds.
The total time of flight, which includes both the upward and downward motions, is twice the time to reach the highest point:
t_total = 2 * t_up ≈ 2 * 1.53 ≈ 3.06 seconds.
Since t = 3.5 seconds is greater than the total time of flight, the ball is in the downward motion at this time.
Therefore, at t = 3.5 seconds, the ball is falling.