Final answer:
The maximum height is 7.35 m, the time of flight is 2.45 s, and the speed when the ball hits the ground is approximately 20.21 m/s. These are calculated using the initial velocities and kinematic equations.
Step-by-step explanation:
The problem given is related to projectile motion, a topic commonly covered in Physics. To solve for the maximum height and time of flight of the ball, we can use the kinematic equations for projectile motion. Given the initial velocities, we can proceed with the following steps:
Maximum Height (c)
To determine the maximum height, we use the vertical motion formula:
h_max = (v_initial_vertical)^2 / (2 * g), where g is the acceleration due to gravity (approximately 9.81 m/s^2). Plugging in the initial vertical velocity of 12 m/s, we get:
h_max = (12 m/s)^2 / (2 * 9.81 m/s^2) = 7.35 m.
Time of Flight (b)
Since the vertical component of initial velocity is 12 m/s, and gravity acts downwards at 9.81 m/s^2, we can use the formula:
t_flight = (2 * v_initial_vertical) / g
This calculation yields:
t_flight = (2 * 12 m/s) / 9.81 m/s^2 = 2.45 s.
Final Speed (a)
For the final speed when the ball hits the ground, we must consider both horizontal and vertical components. The horizontal component remains constant at 16 m/s due to no horizontal acceleration. For the vertical speed, we use:
v_final_vertical = g * t_flight/2
From this, we find:
v_final_vertical = 9.81 m/s^2 * 1.22 s = 11.97 m/s. The final speed is the vector sum of the horizontal and vertical components:
v_final = sqrt((16 m/s)^2 + (11.97 m/s)^2) \≈ 20.21 m/s.