To determine the time the ball remains in the air, we need to consider its vertical motion. The initial vertical speed is given as 3 m/s, and we can assume the ball is launched vertically upwards, experiencing only the force of gravity.
The vertical motion can be described using the equation:
h = h0 + v0t - (1/2)gt^2
Where: h is the vertical displacement at time t h0 is the initial vertical position (assumed to be 0 in this case, since we are measuring from the ground) v0 is the initial vertical velocity (3 m/s upwards) g is the acceleration due to gravity (approximately 9.8 m/s^2) t is the time elapsed
At the highest point of its trajectory, the ball will have a vertical velocity of 0 m/s. So, we can set v = 0 in the equation above and solve for t:
0 = 3 - 9.8t 9.8t = 3 t = 3 / 9.8 t ≈ 0.306 seconds
Thus, the ball will remain in the air for approximately 0.306 seconds.
To calculate the range of the ball, we can consider its horizontal motion. The horizontal speed remains constant throughout the motion, so we can use the equation:
range = horizontal speed * time
range = 6 m/s * 0.306 s range ≈ 1.836 meters
Therefore, the range of the ball will be approximately 1.836 meters.