Question

A ball is thrown straight upward with an initial speed of 19.5 m/s and falls back to the same height. Determine the time of flight and the maximum height reached.

Answer 1

The ball thrown upward with initial speed of 19.5 m/s has a time of flight of approximately 3.98 seconds and reaches a maximum height of about 19.6 meters.

Step-by-step explanation:

To determine the time of flight and maximum height of a ball thrown straight upward with an initial speed of 19.5 m/s, we can use the equations of motion for uniformly accelerated motion. Assuming the acceleration due to gravity (g) is 9.81 m/s2 downward, the time (t) it takes for the ball to reach the maximum height where its velocity becomes zero can be found with the equation v = u +
`v = u + at`, where v is the final velocity (0 m/s), u is the initial velocity (19.5 m/s), and a is the acceleration (-9.81 m/s2). Solving for t gives us t = 19.5 / 9.81, which is approximately 1.99 seconds to reach the maximum height. However, since the ball takes the same amount of time to come back down, the total time of flight is twice this, leading to approximately 3.98 seconds.

The maximum height can be found with the equation
`s = ut + (1/2)at2,`where s is the displacement. Using the value of t determined for the ascent, the calculation gives us s = 19.5 * 1.99 + (1/2) * (-9.81) * (1.99)2, resulting in a maximum height of approximately 19.6 meters.