Answer:
2.22m/s to the left
Explanation:
Given parameters:
Initial position = +125m
Final position = -75m
Motion time = 90s
Unknown:
Velocity of the steel ball = ?
Solution:
The velocity of the steel ball is given as the displacement divided by the time;
Velocity = 
The net displacement of the ball = 125- (-75) = 200m to the left
Input the parameters and solve for the velocity;
Velocity =
= 2.22m/s to the left
Answer:
271.095 m
Explanation:
✓ Let speed of sound in air that was given as (343 m/s) be represented as (Vi)
✓( speed of sound in concrete that was given as (3000 m/s ) be debited as (Vc)
✓ Let the distance travelled by the sound = s
✓duration of Time that exist between heard of sounds = 0.70s
But we know that
Time = (Distance / Speed)
✓Time it takes the sound to travel through air= s/vi = s/343
✓Time it takes the sound to travel through concrete= s/vc = s/3000
✓ (s/343) - (s/3000) = 0.70
Finding LCM and simplify
[(3000s - 343s)]/1029000 = 0.70
2657s /1029000 = 0.70
Making " s" subject of the formula
s= (1029000 × 0.70)/2657
s=720300/ 2657
s= 271.095 m
Hence, The impact took place at a distance of 271.095 m away from the person.