I believe that the correct answer you are looking for is the distance traveled
The initial velocity of the ball is 55.125 m/s.
<h3>Initial velocity of the ball</h3>
The initial velocity of the ball is calculated as follows;
During upward motion
h = vi - ¹/₂gt²
h = vi - 0.5(9.8)(3²)
h = vi - 44.1 ----------------- (1)
During downward motion
h = vi + ¹/₂gt²
h = 0 + 0.5(9.8)(1.5)²
h = 11.025 ----------- (2)
solve (1) and (2) together, to determine the initial velocity of the ball
11.025 = vi - 44.1
vi = 11.025 + 44.1
vi = 55.125 m/s
Thus, the initial velocity of the ball is 55.125 m/s.
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v = v₀ + at
v = final speed, v₀ = initial speed, a = acceleration, t = elapsed time
Given values:
v₀ = 0m/s (starts from rest), a = 9.81m/s², t = 3s
Plug in and solve for v:
v = 0 + 9.81(3)
v = 29.4m/s
A car with a velocity of 22 m/s is accelerated at a rate of 1.6 
for 6.8s has the final velocity t be 32.88 m/s.
The acceleration means the amount of velocity changing per unit time.
The given data:
initial velocity, u = 22 m/s
time, t = 6.8 s
acceleration, a = 1.6
We will be using the equation of motion:
v = u + at
The final velocity become 32.88 m/s.
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Answer:
1. 571.43m/s
2. 142.9m and 342.9m
Explanation:
1.Take the difference in time.
1.2-0.7=0.7 seconds
Take the distance between them and divide with differnce in time.
400÷0.7=571.43 seconds.
2.Take the time of the two men and divide by two.
0.5÷2= 0.25 secs
1.2÷2= 0.6 secs
multiply each with the velocity.
0.25×571.43=142.9m
0.6×571.43=342.9m
