A ball is thrown vertically downward with a velocity of 12m/s How far has the ball fallen 1 s later?. 2s later?
To determine how far the ball has fallen after a certain time, we can use the equation of motion for an object in free fall:
distance = initial velocity × time + 0.5 × acceleration × time^2
Given that the ball is thrown vertically downward, we can assume the acceleration due to gravity (g) is 9.8 m/s^2 (approximately).
1. After 1 second:
Let's calculate the distance traveled by the ball 1 second later:
distance = (12 m/s) × (1 s) + 0.5 × (9.8 m/s^2) × (1 s)^2
distance = 12 m/s + 0.5 × 9.8 m/s^2 × 1 s
distance = 12 m + 0.5 × 9.8 m/s^2
distance = 12 m + 4.9 m
distance = 16.9 meters
2. After 2 seconds:
Now, let's calculate the distance traveled by the ball after 2 seconds:
distance = (12 m/s) × (2 s) + 0.5 × (9.8 m/s^2) × (2 s)^2
distance = 12 m/s × 2 s + 0.5 × 9.8 m/s^2 × (2 s)^2
distance = 24 m + 0.5 × 9.8 m/s^2 × 4 s^2
distance = 24 m + 0.5 × 9.8 m/s^2 × 16 s^2
distance = 24 m + 0.5 × 9.8 m/s^2 × 16 s^2
distance = 24 m + 0.5 × 9.8 m/s^2 × 16 s^2
distance = 24 m + 0.5 × 9.8 m/s^2 × 16 s^2
distance = 24 m + 78.4 m
distance = 102.4 meters
Therefore, the ball has fallen 16.9 meters after 1 second and 102.4 meters after 2 seconds.