To determine the velocity of the ball at different positions during its fall, we can use the equations of motion under constant acceleration. In this case, the acceleration is due to gravity, and we can assume no other significant forces such as air resistance.
Let's calculate the velocity of the ball at two different positions: when it has covered 20m and when it hits the ground.
- Velocity when the ball has covered 20m: We can use the equation of motion:
v^2 = u^2 + 2as
where: v = final velocity u = initial velocity (which is zero since the ball is dropped) a = acceleration due to gravity (approximately 9.8 m/s²) s = distance covered (20m in this case)
Rearranging the equation:
v^2 = 0 + 2 * 9.8 m/s² * 20m v^2 = 2 * 9.8 m/s² * 20m v^2 = 392 m²/s²
Taking the square root of both sides:
v = √(392 m²/s²) v ≈ 19.8 m/s (rounded to one decimal place)
Therefore, when the ball has covered 20m, its velocity is approximately 19.8 m/s.
- Velocity when the ball hits the ground: To determine the velocity when the ball hits the ground, we can use the equation of motion:
v^2 = u^2 + 2as
where: v = final velocity u = initial velocity (zero) a = acceleration due to gravity (-9.8 m/s², negative because it acts opposite to the direction of motion) s = distance covered (40m, which is the height of the tower)
Rearranging the equation:
v^2 = 0 + 2 * (-9.8 m/s²) * 40m v^2 = -2 * 9.8 m/s² * 40m v^2 = -2 * 9.8 m/s² * 40m v^2 = -784 m²/s²
Taking the square root of both sides:
v = √(-784 m²/s²) v ≈ -28 m/s (rounded to one decimal place)
Therefore, when the ball hits the ground, its velocity is approximately -28 m/s. The negative sign indicates that the velocity is directed downward.