Question

Julia throws a ball vertically upward from the ground with a speed of 5.89 m/s. Andrew catches it when it us on its way down at a height of 1.27 m from the ground. After how much time does Andrew catch the ball? (HINT: Find the time it takes to go up and add it to the time it takes to come down)

Answer 1

The ball takes approximately 0.6 seconds to go up and 0.6 seconds to come back down. Therefore, Andrew catches the ball after a total of about 1.2 seconds.

Step-by-step explanation:

First, let's calculate the time it takes for the ball to go up. We can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the ball is thrown vertically upward, the acceleration is -9.8 m/s² (negative because it is directed downward). The final velocity is 0 m/s because the ball reaches its maximum height and then starts to fall back down. The initial velocity is 5.89 m/s. Plugging these values into the formula, we have 0 = 5.89 - 9.8t. Solving for t, we find t ≈ 0.6 s.

Next, let's calculate the time it takes for the ball to come back down. We can use the same formula, but this time the initial velocity is 0 m/s because the ball is at rest when it starts falling. The final velocity is again 0 m/s because the ball is caught. Plugging in these values, we have 0 = 0 - 9.8t. Solving for t, we find t ≈ 0.6 s.

Finally, we add the time it takes to go up and the time it takes to come down to get the total time. t(total) = t(up) + t(down) = 0.6 + 0.6 = 1.2 s.