A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un
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Answer:
Velocity of the ball after 3.04 (s) = 29.79 (m/s)
Explanation:
From the free fall movement we have the following formulas: and
, First we need to find the height to time iqual to 3.04 s using the formula:
and remember that golf ball was released from the rest (Vo= 0 (m/s)) so
, we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using
solving for Vf, we get:
replacing the values given
, so we get: Vf = 29.79 (m/s).
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
- Distance, S = 110 meters
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
Substituting the values into the formula, we have;
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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